ГДЗ_7 кл_2 том_рус

aaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaa

ÁÁÊ ÿ721
Ñ89

Íàçâà «Ãîòîâ³ äîìàøí³ çàâäàííÿ»®
º çàðåºñòðîâàíîþ òîâàðíîþ ìàðêîþ

Ñâ³äîöòâî íà çíàê äëÿ òîâàð³â
³ ïîñëóã ¹ 32091 â³ä 16.06.2003

Îõîðîíÿºòüñÿ Çàêîíîì Óêðà¿íè ïðî àâòîðñüêå ïðàâî.
Ïåðåäðóê äàíîãî ïîñ³áíèêà àáî áóäü-ÿêî¿ éîãî ÷àñòèíè

çàáîðîíÿºòüñÿ áåç ïèñüìîâîãî äîçâîëó âèäàâíèöòâà.
Áóäü-ÿê³ ñïðîáè ïîðóøåííÿ çàêîíó ïåðåñë³äóâàòèìóòüñÿ

ó ñóäîâîìó ïîðÿäêó.

Ñ89 ÑÓÏÅÐ ÃÄÇ. Ãîòîâ³ äîìàøí³ çàâäàííÿ. 7 êëàñ.
Ñ89 Ðîçâ’ÿçàííÿ âïðàâ òà çàâäàíü äî óñ³õ øê³ëüíèõ ï³ä-
ðó÷íèê³â. Êí. 2. — ÒÎÐѲÍà ÏËÞÑ, 2011. — 752 ñ.

ISBN 978-617-03-0015-7.

ßê³ñíà ³ íàéïîâí³øà çá³ðêà äëÿ áàòüê³â! Ïîñ³áíèê ì³ñòèòü
ðîçâ’ÿçàííÿ âïðàâ ³ çàâäàíü äî óñ³õ ï³äðó÷íèê³â, ùî â³äïîâ³äàþòü
ïåðåë³êó ï³äðó÷íèê³â ³ íàâ÷àëüíèõ ïîñ³áíèê³â, ðåêîìåíäîâàíèõ
̳í³ñòåðñòâîì îñâ³òè ³ íàóêè Óêðà¿íè. Ó âèäàíí³ çàïðîïîíîâàí³
ðîçâ’ÿçàííÿ çàâäàíü ç òàêèõ ïðåäìåò³â: àëãåáðà (5 ï³äðó÷íèê³â),
ãåîìåòð³ÿ (5 ï³äðó÷íèê³â), ô³çèêà (3 ï³äðó÷íèêè), õ³ì³ÿ (3 ï³äðó÷-
íèêè), á³îëîã³ÿ (3 ï³äðó÷íèêè), óêðà¿íñüêà ìîâà (3 ï³äðó÷íèêè),
ðîñ³éñüêà ìîâà (2 ï³äðó÷íèêè), àíãë³éñüêà ìîâà (3 ï³äðó÷íèêè),
í³ìåöüêà ìîâà.

Äîïîìîæ³òü äèòèí³ — ïîÿñí³òü íåçðîçóì³ëå!

ÑÓÏÅÐ ÃÄÇ. Ãîòîâûå äîìàøíèå çàäàíèÿ. 7 êëàññ.
Ðåøåíèå çàäàíèé è óïðàæíåíèé êî âñåì øêîëüíûì
ó÷åáíèêàì. Êí. 2. — ÒÎÐÑÈÍà ÏËÞÑ, 2011. — 752 ñ.

ISBN 978-617-03-0015-7.

Êà÷åñòâåííûé è ñàìûé ïîëíûé ñáîðíèê äëÿ ðîäèòåëåé! Ïîñîáèå
ñîäåðæèò ðåøåíèÿ óïðàæíåíèé è çàäàíèé êî âñåì ó÷åáíèêàì,
ñîîòâåòñòâóþùèì ïåðå÷íþ ó÷åáíèêîâ è ó÷åáíûõ ïîñîáèé, ðåêîìåí-
äîâàííûõ Ìèíèñòåðñòâîì îáðàçîâàíèÿ è íàóêè Óêðàèíû. Â ñáîðíèêå
ïðåäëîæåíû ðåøåíèÿ çàäàíèé ïî ñëåäóþùèì ïðåäìåòàì: àëãåáðà
(5 ó÷åáíèêîâ), ãåîìåòðèÿ (5 ó÷åáíèêîâ), ôèçèêà (3 ó÷åáíèêà), õèìèÿ
(3 ó÷åáíèêà), áèîëîãèÿ (3 ó÷åáíèêà), óêðàèíñêèé ÿçûê (3 ó÷åáíè-
êà), ðóññêèé ÿçûê (3 ó÷åáíèêà), àíãëèéñêèé ÿçûê (3 ó÷åáíèêà),
íåìåöêèé ÿçûê.

Ïîìîãèòå ðåáåíêó — îáúÿñíèòå íåïîíÿòíîå!
ÁÁÊ ÿ721

ISBN 978-617-03-0015-7 © ÔÎÏ Øàï³ðî Ì. Â., ìàêåò, 2011

 êàæäîé èç êíèã ñåðèè «ÑÓÏÅÐ ÃÄÇ. Ãîòîâûå
äîìàøíèå çàäàíèÿ» ïðåäñòàâëåíû ðåøåíèÿ âñåõ äî-
ìàøíèõ çàäàíèé è ñàìîñòîÿòåëüíûõ ðàáîò êî âñåì
îñíîâíûì ó÷åáíèêàì ïî âñåì ïðåäìåòàì.

Êíèãà ïðåäíàçíà÷åíà â ïåðâóþ î÷åðåäü òåì ó÷åíè-
êàì, êîòîðûå íå ñòîëüêî ñòðåìÿòñÿ ñïèñûâàòü, ñêîëü-
êî íóæäàþòñÿ â ïîñîáèè, ñ êîòîðûì ìîæíî ñâåðèòü
ñîáñòâåííûå ðåøåíèÿ è ðåçóëüòàòû, à òàêæå ñ åãî ïî-
ìîùüþ ïîíÿòü õîä ðåøåíèÿ ñëîæíûõ çàäàíèé. Ñåðèÿ
«ÑÓÏÅÐ ÃÄÇ. Ãîòîâûå äîìàøíèå çàäàíèÿ» áóäåò ïîëåç-
íîé òàêæå ðîäèòåëÿì, êîòîðûå õîòÿò ïîìî÷ü äåòÿì, íî
óñïåëè îñíîâàòåëüíî ïîäçàáûòü øêîëüíóþ ïðîãðàììó
è íå ìîãóò ðåøèòü çàäà÷è áåç ïîñòîðîííåé ïî-
ìîùè. Äàæå ó÷èòåëþ, ïðè÷åì ñàìîìó îïûòíîìó
è çíàþùåìó, äàííîå èçäàíèå ìîæåò ïðèíåñòè îùó-
òèìóþ ïîëüçó, òàê êàê ðàçíîîáðàçèå ïîäõîäîâ
ê ðåøåíèþ çàäà÷, ïðåäëîæåííûõ â êíèãå, ìîæíî èñ-
ïîëüçîâàòü äëÿ òîãî, ÷òîáû ñòèìóëèðîâàòü ó÷åíèêîâ
ê èçîáðåòåíèþ íîâûõ ïóòåé ðåøåíèÿ.

Æåëàåì óñïåõîâ!

ÑÎÄÅÐÆÀÍÈÅ

Ðåøåíèå çàäàíèé è óïðàæíåíèé ê ó÷åáíèêó
«ÀËÃÅÁÐÀ» Ã. Ï. Áåâçà, Â. Ã. Áåâç ..........................................................................5

Ðåøåíèå çàäàíèé è óïðàæíåíèé ê ó÷åáíèêó
«ÀËÃÅÁÐÀ» À. Ñ. Èñòåðà...................................................................................... 153

Ðåøåíèå çàäàíèé è óïðàæíåíèé ê ó÷åáíèêó
«ÀËÃÅÁÐÀ» Â. Ð. Êðàâ÷óêà, Ã. Ì. ßí÷åíêî........................................................279

Ðåøåíèå çàäàíèé è óïðàæíåíèé ê ó÷åáíèêó
«ÀËÃÅÁÐÀ» À. Ã. Ìåðçëÿêà, Â. Á. Ïîëîíñêîãî, Ì. Ñ. ßêèðà ...........................395

Ðåøåíèå çàäàíèé è óïðàæíåíèé ê ñáîðíèêó
«ÀËÃÅÁÐÀ. Òåìàòè÷åñêîå îöåíèâàíèå» À. Ã. Ìåðçëÿêà, Â. Á. Ïîëîíñêîãî,
Å. Ì. Ðàáèíîâè÷à, Ì. Ñ. ßêèðà..............................................................................479

Ðåøåíèå çàäàíèé è óïðàæíåíèé ê ó÷åáíèêó
«ÕÈÌÈß» Í. Í. Áóðèíñêîé ...................................................................................539

Ðåøåíèå çàäàíèé è óïðàæíåíèé ê ó÷åáíèêó
«ÕÈÌÈß» À. À. Ëàøåâñêîé ................................................................................. 551

Ðåøåíèå çàäàíèé è óïðàæíåíèé ê ó÷åáíèêó
«ÕÈÌÈß» Ï. Ï. Ïîïåëÿ ........................................................................................583

Ðåøåíèå çàäàíèé è óïðàæíåíèé ê ó÷åáíèêó
«ÓÊÐÀÈÍÑÊÈÉ ßÇÛÊ» Í. Â. Áîíäàðåíêî, À. Â. ßðìîëþê.............................599

Ðåøåíèå çàäàíèé è óïðàæíåíèé ê ó÷åáíèêó
«ÓÊÐÀÈÍÑÊÈÉ ßÇÛÊ» À. À. Âîðîí, Â. À. Ñîëîïåíêî ....................................633

Ðåøåíèå çàäàíèé è óïðàæíåíèé ê ó÷åáíèêó
«ÓÊÐÀÈÍÑÊÈÉ ßÇÛÊ» Å. Í. Ãîðîøêèíîé,
À. Â. Íèêèòèíîé, Ë. À. Ïîïîâîé ...........................................................................637

Ðåøåíèå çàäàíèé è óïðàæíåíèé ê ó÷åáíèêó
«ÀÍÃËÈÉÑÊÈÉ ßÇÛÊ» Ë. Â. Áèðêóí ................................................................695

Ðåøåíèå çàäàíèé è óïðàæíåíèé ê ó÷åáíèêó
«ÀÍÃËÈÉÑÊÈÉ ßÇÛÊ» Î. Ä. Êàðïüþê .............................................................727

Ðåøåíèå çàäàíèé è óïðàæíåíèé ê ó÷åáíèêó
«ÀÍÃËÈÉÑÊÈÉ ßÇÛÊ» À. Í. Íåñâèò ................................................................733

АЛГЕБРА

Решение заданий и упражнений
к учебнику
Г. П. Бевза, В. Г. Бевз

7932249490666

§ 1. ОБЩИЕ СВЕДЕНИЯ ОБ УРАВНЕНИЯХ

Óðîâåíü À 3x =1− 5
48
8. à) 25 + x = 37
3x = 3; x = 1.
x = 37 – 25 = 12; 48 2
á) x – 12 = 23
x = 12 + 23 12. à) 2x + 35 — ëåâàÿ ÷àñòü óðàâíåíèÿ,
x = 35;
â) 24 – x = 18 24 — ïðàâàÿ;
x = 24 – 18 = 6;
ã) 3,7 – x = 1,9 â) 34z – 15 — ëåâàÿ; 28z + 3 — ïðàâàÿ.
x = 3,7 – 1,9
x = 1,8; 13. â) 3 ⋅ 7 + 11 = 32;

ä) 1 = 2 + x á) 2 ⋅ 8 – 9 = 15 – 8
3
14. à) x – 2 = 3 x èìååò ðåøåíèå x = –1,
x =1− 2
3 ò. ê. –1 – 2 = 3 ⋅ (–1); –3 ≡ –3;

x = 1; á) 8z – 5 = 5z èìååò ðåøåíèå z = 5 ,
3 3

å) 13 = 74 – x ò. ê. 8 ⋅ 5 − 5 = 5 ⋅ 5 ; 5 ⋅ ⎛ 8 − 1⎠⎟⎞ = 5⋅5.
33 ⎝⎜ 3 3
x = 74 – 13 15. à) x(x –3) = 0
x – 61.
x = 0; 0 ⋅ (0 – 3) = 0;
9. à) 6x = 30
x = 3; 3 ⋅ ( 3 – 3) = 0;
x = 5;
á) 5y = 0 á) z ⋅ (z – 2)(z + 3) = 0
y = 0; z = 0; 0 ⋅ (0 – 2)(0 + 3) = 0
â) 4z = –8 z = 2; 2 ⋅ (2 – 2)(2 + 3) = 0
z = –2;
АЛГЕБРА (к учебнику Г. П. Бевза, В. Г. Бевз) ã) 2x + 3 = 19 z = –3; –3 ⋅ (–3 –2)(–3 + 3) = 0.
2x = 16
x = 8; 16. à) x ⋅ (3x – 1) = 0
ä) 3y – 4 = 1
3y = 5 x=0

y = 5; á) (2n + 2) ⋅ n
3 n=0
n = –1.
å) 1 – 3x = 25
17. à) x – (3 – 2x) = 9
3x = 1 – 25
3x = –24; x – 3 + 2x = 9
x = –8.
3x – 3 = 9
10. à) 2 x = 5
3x = 12
3
x = 5 : 2 = 5 ⋅ 3 = 15 = 7,5; x = 4;

52 2 á) 8 – (3x – 2) = 13
á) − 5 y = 1
8 – 3x + 2 = 13
7
y = −7; 3x + 10 = 13

5 –3x = 3
â) 1 − 3 x = 5
x = –1;
48
â) 3 ⋅ (x – 2) = 27

3x – 6 = 27

3x = 33

x = 11

18. à) 2 ⋅ (x – 3) = 36;

x – 3 = 18

x = 21;

á) 4 ⋅ (5 – x) = 12;
5–x=3
x = 2;

6 â) 0,1 ⋅ (x + 1) = 1

x + 1 = 10 x = 7; АЛГЕБРА (к учебнику Г. П. Бевза, В. Г. Бевз)
x = 9. â) 8z – (5 – 3z) = 17;
8z – 5 + 3z = 17 7
19. à) 3 ⋅ (x + 5) = 27; 11z = 22
z = 2;
x+5=9 ã) 12y + (5 – 2y) = –15
x = 4; 12y + 5 – 2y = –15
á) 5 ⋅ (x – 3) = 15; 10y = –20
x–3=3 x = –2.
x = 6;
â) 8 ⋅ (3 – x) = 40 23. à) 2x – (x – 3) = 20;
3–x=5
x = –2. 2x – x + 3 = 20
x = 17;
20. à) 4 ⋅ (x + 1) + 11 = 31 á) 5 – (4y – y) = 10;
5 – 3y = 10
4(x + 1) = 20
x+1=5 3y = –5; y = − 5 ;
x = 4; 3
á) 16 + 3 ⋅ (z – 2) = 1;
3 ⋅ (z – 2) = –15 â) 4z – (17 + 3z) = 2;
z – 2 = –5 4z – 17 – 3z = 2
z = –3; z = 19;
â) 5 ⋅ (y – 3) – 12 = 73; ã) 17y + (8 – 15y) = 4
5(y – 3) = 85 17y + 8 – 15y = 4
y – 3 = 17 2y + 8 = 4; 2y = – 4; y = –2.
y = 20;
ã) 47 + 2(x + 4) = 7 25. x — çàäóìàííîå ÷èñëî. 3x + 18 = 63;
2 ⋅ (x + 4) =–40
x + 4 = –20 3x = 63 – 18; 3x = 45; x = 15.
x = –24.
26. x — çàäóìàííîå ÷èñëî. 7x – 16 =
21. à) 5 ⋅ (2x – 3) = 50
= 33; 7x = 49; x = 7.
2x – 3 = 10
2x = 13 Óðîâåíü Á
x = 6,5;
á) 37 ⋅ (8x – 23) = 37; 27. à) 2 ⋅ (6 − 9x) = 15;
8x – 23 = 1
8x = 24 3
x = 3; 6 − 9x = 15 : 2
â) 52 ⋅ (17 – 8x) = 52;
17 – 8x = 1 3
8x = 16 6 − 9x = 45
x = 2;
ã) 84 ⋅ (37 – 17z) = 168; 2
37 – 17 ⋅ z = 2 9x = 6 – 22,5
17z = 35
9x = –16,5
z = 35 = 2 1 .
17 17 x = − 11 ;
6
22. à) 3x + (7 – x) = 10;
á) 3 ⋅ (12 − x) = 3 ;
3x + 7 – x = 10
42
2x = 3 12 − x = 3 : 3
x = 3;
24
2 12 – x = 2
á) 2x – (3 – x) = 18;
2x – 3 + x = 18 x = 10;
3x = 21
â) 2 ⋅ (8 − 5x) = 1 ;

55
8 − 5x = 1 : 2

55
8 – 5x = 0,5

5x = 7,5

x =1,5.

АЛГЕБРА (к учебнику Г. П. Бевза, В. Г. Бевз) 28. à) (3x – 2) : 2 = 18; â) 7 : 4 = 5x : 3;
5x ⋅ 4 = 7 ⋅ 3
8 3x – 2 = 36 x = 7 ⋅ 3 = 21

3x = 38 5 ⋅ 4 20
x = 38 = 12 2 ; x=1 1 .

33 20
á) (5x – 3) : 3 = 9; 5x – 3 = 27; 5x = 30; x = 6;
32. â) 2 : m = m : 8;
â) (41 – x) : 9 = 4
m2 = 2 ⋅ 8
41 – x = 36 m = 16
m = ±4;
x = 5. á) x : 10 = 0,1 : x;
x2 = 1
29. à) 2 ⋅ (3x – 2) + 4 = 30; x = ±1;
â) 2n : 9 = 2 : n
2 ⋅ (3x – 2) = 26 2n2 = 18
n2 = 9
3x – 2 = 13 n = ±3.

3x = 15 33. à) x + 5 = 12;

x = 5; x =7
x = ±7;
á) 3 ⋅ (2 – x) + 25 = 28; á) x − 8 = −3;

3 ⋅ (2 – x) = 3 x =8−3

2–x=1 x =5
x = ±5;
x=1 â) 2 x + 3 = 25;

â) 2 ⋅ (x − 4) + 1 = 5; 2 x = 22

33 x = 11
x = ±11;
2 ⋅ (x − 4) = 4 2 Ïðîâåðêà:
1) x = 11; 2 ⋅ 11 + 3 = 25
33 22 + 3 = 25
x − 4 = 4 2 : 2 = 14 ⋅ 3 = 7; x = 11; 25 = 25;
2) x = –11; −11 = 11, äàëåå êàê 1).
3 3 3⋅2
34. à) x + 4 = 0
ã) 3 ⋅ (1 − 2x) + 1 = −2;
x+4=0
77 x = –4;
3 ⋅ (1 – 2x) + 1 = –14; á) x − 2 = 12
x – 2 = ±12
3(1 – 2x) = –15; x = 2 ± 12
x1 = 14; x2 = –10;
1 – 2x = –5; 2x = 6; x = 3. â) x − 1 + 7 = 3

30. à) 6x : 8 = 3 : 2; x − 1 = −4
∅ (ðåøåíèé íåò).
6x ⋅ 2 = 8 ⋅ 3
x = 8 ⋅ 3 = 2; 35. à) 2x − 3 = 5

6⋅2 2x – 3 = ±5
á) 5 : (2x) = 3 : 18; 2x = 3 ± 5
2x = 8; x1 = 4
2x ⋅ 3 = 5 ⋅ 18 2x = –2; x2 = –1;
x = 5 ⋅ 18 = 15;

2⋅3
â) 3x ⋅ 4 = 1 ⋅ 12
x = 1 ⋅ 12 = 1.

3⋅4

31. à) (x – 5) : 2 = 3 : 4;

(x – 5) ⋅ 4 = 2 ⋅ 3
x −5 = 2⋅3

4
x−5= 3

2
x = 61;

2
á) 5 : (c – 3) = 2 : 3;

(c – 3) ⋅ 2 = 5 ⋅ 3

c − 3 = 5 ⋅ 3 c − 3 = 15 = 7,5
22

c = 10,5;

á) 2x − 3 = 5; 40.

2x = 8 x
2x = ±8
x1 = 4; x2 = –4; 2x
â) 2 ⋅ x − 3 = 5;
P = 50 ñì; (x + 2x) ⋅ 2 = 50
x − 3 = 2,5; x – 3 = 2,5
x1 = 3 + 2,5 = 5,5 3x = 25;
x2 = 3 – 2,5 = 0,5.
x = 8 1 ; 8 1 ⋅ 2 = 25 ⋅ 2 = 50 = 16 2 .
36. à) 3ax + 96 = 0. Åñëè x = –8, òî 33 33 3

3a ⋅ (–8) + 96 = 0; a = −96 = 12 = 4; Øèðèíà ïðÿìîóãîëüíèêà 8 1 ñì, äëè-
−3 ⋅ 8 3 3
íà 16 2 ñì.
á) 1 − a ⋅ x = − 1 . Óðàâíåíèå èìååò êîðåíü 3
42
41.
x = 2, åñëè 1 − a ⋅ 2 = − 1 ; a ⋅ 1 = 3 ; a = 3;
4 222 x 2x

â) 4(a – 3) ⋅ x = 72 èìååò êîðåíü x = 6 P2 – P1 = 50 ñì
ïðè 4 ⋅ (a – 3) ⋅ 6 = 72; a − 3 = 72 ; a – P2 = 2 x ⋅ 4 = 8x
P1 = x ⋅ 4
24 P2 – P1 = 8x – 4x = 4x; 4x = 50; x = 12,5
3 = 3; a = 6. 12,5 ⋅ 2 = 25.

37. à) 2 ⋅ (x + 3) = 36 è x : 3 + 2m = 19. Îòâåò: ñòîðîíà âòîðîãî êâàäðàòà 25 ñì.
42. Äåäó x ëåò. x + x + x = 200; 2,5x =
x + 3 = 18 2 АЛГЕБРА (к учебнику Г. П. Бевза, В. Г. Бевз)
x = 15 ïîäñòàâèì âî âòîðîå óðàâíåíèå: = 200; x = 80.
15 : 3 + 2m = 19; 2m = 19 – 5; m = 7;
á) (8 – x) ⋅ 7 = 28 è 5 ⋅ (2x – 3m) = 0 Îòâåò: 80 ëåò.
8–x=4
x = 4 ïîäñòàâèì âî âòîðîå óðàâíåíèå: 43. ×åðåç x ëåò îòöó áóäåò (34 +
5 ⋅ (2 ⋅ 4 – 3m) = 0; 8 – 3m = = 0; m = 8 ;
+ x) ëåò, äî÷êå — (12 + x) ëåò, òî åñòü
3
â) (x : 3 + 8) ⋅ 2m = 48 è (3x – 2) : 2 = 17 (12 + x) ⋅ 2 = 34 + x; 24 + 2x = 34 +
3x – 2 = 34; 3x = 36; x = 12 ïîäñòàâèì
â ïåðâîå óðàâíåíèå + x; x = 10.
(12 : 3 + 8) ⋅ 2m = 48; 12 ⋅ 2 ⋅ m = 48;
m = 2. Îòâåò: ÷åðåç 10 ëåò äî÷êà áóäåò ìëàä-

38. à) x2 = k. Ïðè k < 0. øå îòöà âäâîå.

á) x + k = 0; x = −k. Ïðè k > 0. 44. Îòâåò: 315 áûêîâ.
â) k + 2x = 2(x – 3); k + 2x = 2x – 6. Ïðè
k ≠ 6 êîðíåé íåò. Óïðàæíåíèÿ äëÿ ïîâòîðåíèÿ

39. 45. à) 3,7 – 1,2 : 0,4 = 3,7 – 3 = 0,7;

x á) 2,8 + 8,1 : 2,7 = 2,8 + 3 = 5,8;

3x â) (7 – 8,5) : 0,5 = –1,5 : 0,5 = –3;

P = 60 ñì; (x + 3x) ⋅ 2 = 60; 4x = 30; x ã) –4,9 : (2,3 – 1,6) = –4,9 : 0,7 = –7;
= 7,5; 3x = 22,5.
Îòâåò: øèðèíà ïðÿìîóãîëüíèêà 7,5 ñì, ä) 12,1 : 0,11 + 1 : (–0,2) = 110 – 5 = 105;
äëèíà 22,5 ñì. å) 0,23 + 0,32 = 0,008 + 0,09 = 0,098;
æ) (3 – 1,4) : 0,22 = 1,6 : 0,04 = 40;
ç) (–0,4)2 – 1,22 = 0,16 – 1,44 = –1,28. 9

46. à) 2a + 5 = 2 ⋅ 2 + 5 = 9;

á) Åñëè m = 8, òî 2,3 – 3m = 2,3 – 3 ⋅ 8 =

= 2,3 – 24 = –21,7;

â) 2a + 3c =
2 ⋅ 1 + 3 ⋅ 1 = 2 + 3 = 4 + 9 = 13 = 2 1 ;

3 232 6 6 6

ã) 2(a + 3c) = 2 ⋅ ⎛ 1 + 3 ⋅ 1⎞ = 2⋅2+9 = 2 ⋅ 11 = 11 = 3 2 .
⎜⎝ 3 2⎟⎠ 6 6 33

47. 2 : 3 : 5. 2x + 3x + 5x = 300; 10x = 300; x = 30.

×èñëà: 2x = 2 ⋅ 30 = 60; 30x = 3 ⋅ 30 = 90; 5x = 5 ⋅ 30 = 150.

§ 2. РАВНОСИЛЬНЫЕ УРАВНЕНИЯ

10 АЛГЕБРА (к учебнику Г. П. Бевза, В. Г. Бевз) Óðîâåíü À 55. à) 7x + 4 = 9x

52. à) 5 + (x – 4) = 5 è x – 4 = 0 4 = 9x – 7x;
á) 38 – 2n = 2n
x–4=0 38 = 2n + 2n;
â) 1 – 0,5z = 1,5z
x=4 1 = 0,5z + 1,5z.

x=4 56. à) 12y + 3 = y – 7

Âûâîä: óðàâíåíèÿ ðàâíîñèëüíû. 12y – y = – 3 – 7;
11y = –10
á) (x – 5)(x + 1) = 0 è x – 5 = 0 y = − 10 ;

x1 = 5; x2 = –1 11
x=5 á) 5x + 2x + 5 = 4x;
7x – 4x = –5
Âûâîä: óðàâíåíèÿ ðàâíîñèëüíû. 3x = –5
x = −5;
â) (x – 3)(x + 3) = 0 è x = 3
3
x1 = 3 â) 0,7 – 2c = 3c + 1,7;
x1 = 3 –2c – 3c = 1,7 – 0,7
x2 = –3 –5ñ = 1
x2 = –3 c = − 1 = −0, 2.
Âûâîä: óðàâíåíèÿ ðàâíîñèëüíû.
5
ã) (x – 5)(x + 1) = 0 è 2 − x = 3
57. à) 2x – 1 = 3x;
x1 = 5
⎡2 − x = 3 2x – 3x = 1
⎢⎣2 − x = −3 –x = 1
x2 = –1 x = –1;
x1 = –1 á) 5y + 6 = 2y
x2 = 5 5y – 2y = –6
Âûâîä: óðàâíåíèÿ ðàâíîñèëüíû. 3y = –6
y = –2;
53. à) 7x + 8 = 10 è 4x – 10 = –3x – 8; â) 0,8z – 1 = 0,3z
0,8z – 0,3z = 1
á) 38 – 2n = 2n è 19 – 2n = 0; 0,5z = 1
z = 1 : 0,5 = 2;
â) 5x – 2 = 2x – 5 è 3x = –3; ã) 2 + 37t = 40t
37t – 40t = –2
ã) x + 3 = 3 − x è x + 3 = 15 – 5x; –3t = –2
5 t = 2;

ä) 1 (3 − 6x) = 3 è 3 – 6x = 1,5; 3

24

ä) 1 + 3x = x è 2x + 1 = 0.
33

54. à) 12x – 3 = x + 2;

12x – x = 3 + 2;

á) 15z + 8 = 2z;

15z – 2z = –8;

â) 1 m − 2 = 3 − 1 m,
23 2

1m + 1m = 3+ 2.
22 3

ä) 1 – 0,5c = 0,5c ã) 5y = − 5 + y; 11 АЛГЕБРА (к учебнику Г. П. Бевза, В. Г. Бевз)
1=c 8
c = 1;
å) 3 + 4,7x = 4,7x; 3 ≠ 0; ∅ 5y − y = − 5 ;
8
58. à) 3(x – 5) = 2x – 7;
4y = − 5
3x – 15 = 2x – 7 8
x = 15 – 7
x = 8; y = −5 :4= − 5 ;
á) 4(l – 0,9) = 1,2 + 2l; 8 32
4l – 3,6 = 1,2 + 2l
4l – 2l = 1,2 + 3,6 ä) 2 x + 8 = 8;
2l = 4,8 3
l = 2,4;
â) 7x – 4(x – 3) = 12; 2x = 0
7x – 4x + 12 = 12 3
3x = 0 x = 0;
x = 0;
ã) 16 – (2 – 5x) = 29; å) −4c = 1 − 1 c
2 – 5x = 16 – 29 77
–5x = –1 – 2
–5x = –15 −4c + 1 c = 1
x = 3. 77

59. à) 1 x = 12 − x; −3 6 c = 1
77
3
1 x + x = 12 c=− 1 .
3 27
4 x = 12
3 60. à) ðàçíîñòü 2x – 3 ðàâíà ñóììå x + 17;
x = 12 ⋅ 3 = 9
2x – 3 = x + 17; x = 20;
4
x = 9; á) x + 37 = 2 ⋅ (x – 15); x + 37 = 2x –
á) 2 y = 9 − y;
– x; x = 67.
3
2y+y =9 Óðîâåíü Á
3
5y = 9 61. à) 1 (2x − 3) = 1;
3
y = 9 : 5 = 9 ⋅ 3 = 27 5
2x – 3 = 5
35 5
y = 5,4; 2x = 8

â) 1 z = 1 + z; x = 4;
63
á) 1 (4 + 3x) = 1 ;
1z−z= 1
63 93
−5z= 1 4 + 3x = 3

63 3x = –1

z = 1 : ⎛ − 5 ⎞ = − 1⋅ 6 = −2 x = −1;
3 ⎝⎜ 6 ⎟⎠ 3⋅5 5 3

z = −2; â) 3 (2 − 3x) = 1 ;
5
77
3(2x – 3x) = 1

6 – 9x = 1

5 = 9x

x = 5.
9

62. à) 2x + 7 = 1;

39
6x + 7 = 9

6x = 2

x = 1;
3

á) 3c − c = 7;
5 10

6c – c = 70

5c = 70 x = 12;
c = 14;
â) n − 3n = −2; å) 1 (7 − 2x) = 3 ⎛ 9x + 4 2⎞ ;
4 ⎜⎝ 3⎠⎟
48 2
2n – 3n = –16;
–n = –16 2 (7 − 2x) = 3 ⎛ 8x + 4 2⎞ ;
n = 16. ⎜⎝ 3⎠⎟

14 – 4x = 24x + 14;

0 = 28x

63. à) 2 + x − 2 = 5 ; x = 0.

93 9 65. à) 2 (x − 2) = 2 ⋅ (5x − 24);

2 + 3(x – 2) = 5 35
3x – 6 = 10x – 48;
3(x – 2) = 3 48 – 6 = 10x – 3x
7x = 42
x–2=1 x = 6;

x = 3; á) 5 (x + 2) = 1 (7x + 12);

á) 3 − y + 5 = −2 2 ; 66
8 2 4 5x + 10 = 7x + 12
3 4 (y + 5) = − 9; –2x = 2
x = –1;
− â) 0,4 ⋅ (6x – 1) = 0,1(12x + 5);
88 4 2,4x – 0,4 = 1,2x + 0,5
1,2x = 0,9
3 – 4 ⋅ (y + 5) = –18 x = 9 = 3;

21 = 4(y + 5) 12 4
ã) 0,5 ⋅ (7x + 8) = 1,5 ⋅ (7x + 8);
y + 5 = 21 = 5 1 3,5x + 4 = 10,5x + 12
44 3,5x – 10,5 = 12 – 4
–7x = 8
y = 1; x = −8;
4
7
â) x − 3 − 2x = 2
39 ä) 5x − 2 = 1 (3x + 4);

12 АЛГЕБРА (к учебнику Г. П. Бевза, В. Г. Бевз) 3x – 9 – 2x = 18 33
15x – 2 = 3x + 4
x = 27. 12x = 6
x = 1;
64. à) 5 ⋅ (0,6m – 2) = 2(m – 3,6);
2
3m – 10 = 2m – 7,2
å) y − 3 ⋅ (2y − 5) = 1 1 − 2y;
m = 10 – 7,2
44
m = 2,8; y − 3 y + 15 = 5 − 2y

á) 3 ⋅ (1,2n + 8) = 4 ⋅ (5 – 0,1n); 2 44

3,6n + 24 = 20 – 0,4n

3,6n + 0,4n = 20 – 24

4n = –4

n = –1;

â) 2 ⋅ (11 – 6x) – 3(7 – 4x) = 1

22 – 12x – 21 + 12x = 1

1=1

x — ëþáîå ÷èñëî.

ã) 7(y + 6) = 4(3y – 5) = 34 y − 3 y + 2y = 5 − 15 ;
2 44
7y + 42 = 12y – 20 – 3

7y – 12y = –42 – 23 1,5 ⋅ y = − 10
4
–5y = –65;

y = 13; y = −5⋅2 = −5.
2⋅3 3
ä) 1 (6 + x) = 2 (2x − 15);
66. à) 2,5 ⋅ (y + 6) = y + 1,5 ⋅ (y – 10);
33
2,5y + 15 = y + 1,5y – 15;
2 + 1 x = 4 x − 10;
33 2,5y – 2,5y = –15 – 15

2 + 10 = 4 x = 1 x 0y = –30
33
∅ — ðåøåíèé íåò.
12 = x

á) 0,75 ⋅ (4 – x) – 0,5x = 5 ⋅ (0,05x + 3); 0,2x = 0,3 ⋅ (425 – x); 0,2x = 127,5 – 13 АЛГЕБРА (к учебнику Г. П. Бевза, В. Г. Бевз)
– 0,3x; 0,5x = 127,5; x = 255;
3 ⋅ (4 − x) − 0,5x = 0,25x + 15; 425 – 255 = 170.
Îòâåò: 255; 170.
4
3 − 3 x − 0,5x − 1 x = 15 74. Áûëî

44 I ñêëàä — 2x
–1,5x = 15 – 3; II ñêëàä — x
–1,5x = 12; x = –8. Ñòàëî:

67. à) 3x + 7 − x − 3 = 5x + 2 ; 6x + 2x + 84
x + 140
42 8 2x + 84 = x + 140; x = 140 – 84 = 56;
+ 14 – 4x + 12 = 5x + 2; ñòàëî íà I ñêëàäå 56 ⋅ 2 + 84 = 196 ò,
íà II ñêëàäå 56 + 140 = 196 ò, òî åñòü
6x – 4x – 5x = 2 – 26; –3x = –24; x = 8; ïîðîâíó.
á) 5x − 4 = 7 − x + 3 + 1 ; 2(5x – 4) = Îòâåò: íà I ñêëàäå 2 ⋅ 56 = 112 ò, íà
II — 56 ò.
3 26
= 3(7 – x) + 3x + 1; 75. I áàê — 2x

10x – 8 = 21 – 3x + 3x + 1; 10 x = 22 + 2x – 17
+ 8; 10õ = 30; x = 3. II áàê — x
x + 17
68. à) 8 − 3y − 1 − 2y = 6y + 17 ; ïîðîâíó 2x – 17 = x + 17;
x = 34; 34 ⋅ 2 = 68.
5 2 10 Îòâåò: â I áàêå 68 ë, âî II áàêå 34 ë
16 – 6y – 5 + 10y = 6y + 16; áåíçèíà.
4y – 6y = 17 – 11;
–2y = 6; y = –3; 76. à) 2(x – 1) = 4 – x è ax = x + a
á) 2 − 4x = 1 − 2x − x + 3 ; ⋅60 12 ⋅ (2 –
2x – 2 = 4 – x
5 34 3x = 6
– 4x) – 5(x + 3); x = 2 ïîäñòàâèì âî âòîðîå óðàâíåíèå,
24 – 48x = 20 – 40x – 5x – 15 èìååì a ⋅ 2 = 2 + a; 2a – a; a = 2.
45x – 48x = 5 – 24; –3x = –21; x = 7. Îòâåò: ïðè a = 2.
á) (1 – a)x = x è x2 = 0. Ïðè x = 0 ïåðâîå
69. 135 – x = (83 – x) ⋅ 3; 135 – x = óðàâíåíèå èìååò âèä:
(1 – a – 1) ⋅ x = 0; –a ⋅ x = 0, òàê ÷òî a ≠ 0.
= 249 – 3x; 3x – y = 249 – 135;
2x = 114; x = 57. 77. 1) 3 (x + a) + 0,5 + 3 = 4 ïðè

70. x + 207 = 4 ⋅ (x + 33); x + 207 = 2

= 4x + 132; 3x = 207 – 132; x = –1; 3 ⋅ (a − 1) − 0,5 + 3 = 4;
3x = 75; x = 25.
Ïðîâåðêà: 1) 25 + 207 = 232; 2
2) 25 + 33 = 58;
3) 232 : 58 = 4, ÷òî îòâå÷àåò óñëîâèþ 3 (a − 1) = 1,5; a – 1 = 1; a = 2.
çàäà÷è.
2
71. Ïðèìåì çà x ïåðâîå ÷èñëî, òîãäà 2) Ïðè x = 5; a = 2 âûðàæåíèå ðàâíî
3 (2 + 5) + 0,5 ⋅ 5 + 3 = 21 + 2,5 + 3 =
âòîðîå ÷èñëî 120 – x;. x = 120 − x ; 22
53 = 10,5 + 5,5 = 16.

3x = 5 ⋅ (120 – 5x); 3x = 600 – 5x; 9x = Îòâåò: 16.

= 600; x = 75. 78 . 1) 9x − m − 4 ⋅ x + 2 = 16
I ÷èñëî 75; II ÷èñëî 120 – 75 = 45.
5
72. x — I ÷èñëî; (x – 12) — II ÷èñëî; ïðè x = 1 2 = 5 .

0,3x = 0,7(x – 12); 0,3x = 0,7x – 8,4;. 33
8,4 = 0,7x – 0,3x; 0,4x = 8,4; x = 21; 9 ⋅ 5 − m − 4 ⋅ 5 + 2 = 16;
21 – 12 = 9.
Îòâåò: 21; 9. 3 53

73. I ÷èñëî — x; II ÷èñëî — 425 – x;

20 % ⋅ x = 30 % ⋅ (425 – x);

15 + 2 − 16 = m − 4 ; m − 4 = 1; = 9 ⋅ ⎛ − 1⎞ − 7 − 4 ⋅ ⎛ − 1⎞ + 2 =
33 ⎜⎝ 6⎟⎠ 5 ⎜⎝ 6⎠⎟

m – 4 = 3; m = 7. − 3 + 3 ⋅ 1 + 2 = 1,5 + 0,1 + 2 = 0,6.
2) Ïðè m = 7 è x = − 1 çíà÷å- 2 5⋅6

6 Îòâåò: 0,6.
íèå âûðàæåíèÿ 9x − m − 4 ⋅ x + 2 =

5

Óïðàæíåíèÿ äëÿ ïîâòîðåíèÿ

80. à) 80 2 512 2
2 256 2
40 2 128 2
20 2 64 2
10 5 32 2
5 1024 = 210;
80 = 24 ⋅ 5; 2 â) 1001 = 7 ⋅ 11 ⋅ 13.
á) 1024

81 ⎛ 4, 3 ⋅ 3 + 11 2 ⋅ 2, 25⎟⎞⎠ : 2,75 = ⎛3 + 58 ⋅ 9⎞ :23 = ⎛3 + 26,1⎟⎠⎞ : 11 = 26, 4 : 11 =2
⎜⎝ 43 5 ⎝⎜ 10 5 ⋅ 4 ⎟⎠ 4 ⎝⎜ 10 44

= 26, 4 : 11 = 132 ⋅ 4 = 9, 6.
4 5 ⋅ 11
Îòâåò: 9,6.

14 АЛГЕБРА (к учебнику Г. П. Бевза, В. Г. Бевз) 82. à) m + n ; á) 2x − 3z .

22

83. 1) 40 – 32 = 8; (8 ⋅ 100) : 32 = 25 %; 2) (8 ⋅ 100) : 40 = 20 %.

§ 3. ЛИНЕЙНЫЕ УРАВНЕНИЯ

Óðîâåíü À â) 3x + 2(x + 7) = 2x;
3x + 2x + 14 = 2x
87. à) 2 – 3x = 5 – 7x; 3x = –14;
ã) 4 ⋅ (2 + x) – x = 3x + 9;
7x – 3x = 5 – 2 8 + 4x – x – 3x = 9
0 ⋅ x = 1 ðåøåíèé íåò;
4x = 3; ä) –c + 31(2 – c) = 32c;
–c + 62 – 31c – 32c = 0
á) 0 = 7x – 5; –64 ⋅ c = –62
èëè 32 ⋅ c = 31;
7x = 5; å) 0,7 = 2 ⋅ (x + 3,5) – 2x;
0,7 = 2x + 7 – 2x
â) 1 x = 6 − 1 x; 0 ⋅ x = 6,3 ðåøåíèé íåò.
32
89. à) 2x – 3 = x + 7;
1 x + 1 x = 6;
32 x = 10
5 x = 6; îäíî ðåøåíèå;
6 á) 3x + 7 = 3x – 9;
ã) x − 4 = 1; ðåøåíèé íåò.
â) 2 ⋅ (3x – 1) = 3(2x + 1)
2 6x – 2 = 6x + 3;
x–4=2 ðåøåíèé íåò.

x = 6.

88. à) 2x + x – 7x + 3 = 8;

–4x = 5;

á) y – 5y = 8 – y;

y – 5y + y = 8

–3y = 8;

90. à) 32x = –16; 94. à) 8(9 – 2x) = 5 ⋅ (2 – 3x);

x= −1 72 – 16x = 10 – 15x
2 –x = –62
x = 62;
á) –15z = 0,5; á) 5 ⋅ (z + 3) = 8(10 – z);
z = − 0,5 = − 1 ; 5z + 15 = 80 – 8z
13z = 65
15 30 z = 5;
â) x + 4x = 5x; â) 2 ⋅ (x – 3) = 3(2x – 1);
2x – 6 = 6x – 3
x — ëþáîå – 3 = 4x
ã) –0,5y = –0,5; y = 1; x = −3;
ä) 6x = 8 + 6x
∅ ðåøåíèé íåò 4
å) x – 4x = 5x; ã) 4 ⋅ (5 – x) = –5x + 2;
–9x = 0; 20 – 4x = –5x + 2
x = 0. x = –18.

91. à) 0 ⋅ x = 35; 95. à) y – 1,08 = 0,2 ⋅ (5 + y);

∅ — óðàâíåíèå íå èìååò ðåøåíèé. y – 1,08 = 1 + 0,2y
á) 0 ⋅ y = 13 – 13; y – 0,2y = 1 + 1,08
0⋅y=0 0,8y = 2,08
y = 2,6;
y — ëþáîå; á) 0,3 ⋅ (1 – c) = c + 0,04;
â) 2x = 3 + 2x 0,3 – 0,3c = c + 0,04
∅ — óðàâíåíèå íå èìååò ðåøåíèé. 0,3 – 0,04 = c + 0,3c
1,3c = 0,26
92. à) 0,5z = 6+ 1 c = 0,2;
z; â) 3 – 5x = 0,3 ⋅ (2x + 1);
3 3 – 5x = 0,6x + 0,3
–5,6x = –2,7
1z− 1z = 6 x = 27 ;
23
56
z = 36; ã) 1 – 3(x – 5) = 7 ⋅ (3 – 2x); 15 АЛГЕБРА (к учебнику Г. П. Бевза, В. Г. Бевз)
1 – 3x + 15 = 21 – 14x
á) 0, 2x + 5 = 1 11x = 5
x; x= 5 .
5
11
1 x − 1 x = −5
55 96. à) − 1 x = 1 x = 3 x;

0 ⋅ x = –5 234
−1x− 3x = 1
∅ óðàâíåíèå íå èìååò ðåøåíèé;
24 3
â) 2 x + 7 = 0, 6x; −5x = 1
3
43
2 x − 3 x = −7 x = − 1 : 5 = −1⋅4 = − 4
35
3 4 3 ⋅ 5 15
1 x = −7; x = –105. x=− 4 ;
15
15
93. à) 4 – 3x = 9(1 – x); á) 2 t − 1 = 0, 6t;

4 – 3x = 8– 8x 35
2t− 6 t = 1
5x = 4 3 10 5
2t− 3t = 1
x = 0,8; 35 5

á) 2 – 5y = 5(1 – 2y);

2 – 5y = 5 – 10y;

5y = 3

y = 0,6;

â) x = 3 ⋅ (x + 1) – 2x;

x = 3x + 3 – 2x

0 ⋅ x = 3; ∅;

ã) 2(5 – 8x) = –4 ⋅ (4x + 3);

10 – 16x = –16x – 12; ∅

1 t=1 10z + 5z = 1 + 2,4
15 5
t = 3; 15z = 3,4
â) −0, 8z + 1 = 4 z;
z = 34 = 17 ;
5 150 75
–0,8z – 0,8z = –1
–1,6z = –1 ã) 2,5x – 1,7 ⋅ (5 – 2x) = 3x;
z = 1 = 10 = 5
2,5x – 8,5 + 3,4x – 3x = 0
1, 6 16 8
z = 5. 2,9x = 8,5

8 x = 85 = 2 27 .
29 29
98. à) 492x + 317 = 923;
101. à) 8 + 3(x – 5) + x = 2 ⋅ (3 + 2x);
492x = 606;
x = 606 : 492 ≈ 1,2317… 8 + 3x – 15 + x = 6 + 4x
x ≈ 1,232
á) 2,38z – 5,87 = 3,41 4x – 7 = 4x + 6
2,38z = 9,28
z = 9,28 : 2,38 = 3,899159… ∅ — óðàâíåíèå íå èìååò ðåøåíèé.
z ≈ 3,89916.
á) z + 2 ⋅ (4 + z) = 3z + 8;
Óðîâåíü Á
z + 8 + 2z = 3z + 8
99. à) 3 ⋅ (x + 4) + 6(11 – x) = 9; 3z = 3z

3x + 12 + 66 – 6x = 9 z — ëþáîå ÷èñëî.
–3x = 9 – 78
– 3x = –69 â) 1 x + 1 (x − 2) = x;
x = 23
á) 8(1 – x) + 5(x – 2) = 2; 24
8 – 8x + 5x – 10 = 2
–3x = 2 + 2 1x+ 1x− 1 −x = 0
–3x = 4 242
x = −4;
−1x = 1
3 42
â) 7(x – 5) – 3(2x – 6) = 10;
7x – 35 – 6x + 18 = 10 x = –2;
x = 10 + 17
x = 27; ã) 1 + 2 ⋅ ⎛ 1 n + 1⎟⎞⎠ = 3n;
ã) 5 ⋅ (3 – 2x) – (12 + 7x) = 0; 2 ⎝⎜ 3
16 АЛГЕБРА (к учебнику Г. П. Бевза, В. Г. Бевз) 15 – 10x – 12 – 7x = 0
–17x = –3 1 + 2 n + 2 − 3n = 0
x= 3 . 23

17 −2 1 n = −2 1 ; n = 5 : 7 = 5 ⋅ 3
3 2 2 3 2⋅7
100. à) 7 ⋅ (4 – t) + 3(t – 5) = 9t;
n = 15 = 1 1 .
28 – 7t + 3t – 15 – 9t = 0 14 14
–13t = –13
t=1 102. à) 1 ⋅ (4x − 5) + 3 ⋅ (2x + 1) = x + 3;
á) 3 ⋅ (x + 1,5) + 2(3 + x) = –5;
3x + 4.5 + 6 + 2x = –5 22
5x = –5 – 10,5 2x – 2,5 + 3x + 1,5 = x + 3
5x = –15,5
x = –3,1; 4x = 4
â) 4z – 1,2 ⋅ (2 – 5z) = 1 – 5z;
4z – 2,4 + 6z = 1 – 5z x = 1;

á) 2 ⋅ (5 − 3x) + 1 ⋅ (2 + 9x) = 2x − 1;

33
10 − 2x + 2 + 3x − 2x = −1
33
–x = –1 – 4; –x = –5

x = 5;

â) 3 ⋅ (6 + 7x) − 2x = 2 ⋅ (4 + 3x) = 3;

55
3 ⋅ (6 + 7x) – 10x = 2(4 – 3x) + 15

18 + 21x – 10x = 8 – 6x + 15

11x – 6x = 23 – 18

5x = 5; x = 1;

ã) 2 + 1 ⋅ (8x + 1) = 5x + 3 ⋅ (4x − 1);

44
2 + 2x + 1 = 5x + 3x − 3

44

2 + 1 + 3 = 6x á) x + 1,5 = 1,5x
44 1,5 = 0,5x
x = 3.
3 = 6x; x = 1 .
2 109. à) m − 14 = 0,2 ⋅ m + 14 ;

103. à) 3 (2x + 3) – 57 – 4x) – 2(5x + 22
m – 14 = 0,2m + 2,8
+ 4) = –2;
6x + 9 – 35 + 20x – 10x – 8 = –2 m – 0,2m = 14 + 2,8
16x = –2 + 34
16x = 32 0,8m = 16,8
x = 2;
á) 8(4 – 3x) + 7(x – 3) + 3(9 + 7x) = 10; m = 21;
32 – 24x + 7x – 21 + 27 + 21x = 10
4x = 10 – 38; 4x = –28; x = –7; á) m + 14 = 1,2 (m − 14);
â) 6(x + 2) + 3(3x + 7) = 4(5 + 4x) – 7;
6x + 12 + 9x + 21 = 20 + 16x – 7 2
15x + 33 = 13 + 16x m + 14 = 2,4(m – 14)
–x = –20; x = 20;
ã) 5(12 – x) – 11(4x – 50 = 9(9 – 5x) – 26; m + 14 = 2,4m – 33,6
60 – 5x – 44x + 55 = 81 – 45x – 26
–49x + 115 = 55 – 45x 47,6 = 1,4m
60 = 4x; x = 15.
m = 476 = 34.
104. I ÷èñëî — (x + 6) ⋅ 5. II ÷èñëî — 14

4x, ÷òî íà 40 ìåíüøå ïåðâîãî 110. I — x
9x = 60 ⋅ 5 — áîëüøå íà 40.
x⋅4 II — 2x
Óðàâíåíèå 5(x + 6) – 40 = 4x
4x = 30 – 40 = 4x III — 3 ⋅ (2x) = 6x
x = 10; 10 + 6 = 16.
IV — 4 ⋅ (6x) = 24x
Îòâåò: 16; 10.
Âìåñòå 132
105. I ÷èñëî — 6x. II ÷èñëî — x
Óðàâíåíèå:
6x – 37,
x + 2x + 6x + 24x = 132
x + 73, ðàâíû
Óðàâíåíèå: 6x – 37 = x + 73 33x = 132; x = 4.
5x = 110; x = 22; 22 ⋅ 6 = 132.
Îòâåò: ïåðâûé ïîæåðòâîâàòåëü äàë 4.
Îòâåò: 132; 22.
111. Ïóñòü ëåòåëî x ãóñåé. Óðàâíåíèå: 17 АЛГЕБРА (к учебнику Г. П. Бевза, В. Г. Бевз)
106. à) I + II = 155
x + x + x + x + 1 = 100; 2 3 x = 99;
I – II = 91 24 4
1) I ÷èñëî = (155 + 91) : 2 = 246 : 2 = 123.
2) II ÷èñëî = (155 – 91) : 2 = 64 : 2 = 32. 11 ⋅ x = 99;
4
Îòâåò: 123; 32.
á) I – II = 47 x = 99 : 11 = 99 ⋅ 4 = 36 ãóñåé ëåòåëî.
I+II = 46 èëè 4 11

2 112. Ïðèìåì âåñ ãîëîâû çà x ã; ïîëî-
I – II = 47
I + II = 92. Ðåøåíèå àíàëîãè÷íî. âèíà òóëîâèùà — (x – 150) ã; âñå òó-

Îòâåò: 69,5 è 22,5. ëîâèùå âåñèò (x + 150) ã. Óðàâíåíèå:

x + 150 = 2 ⋅ (x – 150).

x + 150 = 2x – 400

450 = x.

Ãîëîâà âåñèò 450 ã; òóëîâèùå âå-

ñèò 450 + 150 = 600 ã, õâîñò — 150 ã.

Îòâåò: ðûáà âåñèò 450 + 600 + 150 =

= 1200 ã.

113*. à) (a2 + 3) ⋅ x = 5
a2 + 3 > âñåãäà.

107. x − 1 x = 1 ; 2 x = 1 ; 2x = 14 x = 1 . Òàê, x= 5 — åäèíñòâåííûé êîðåíü
a2 + 3
ïðè ëþáîì a.
3 33 3 2

108. à) x + 15 = 2 ⋅ (x – 15)4 á) (a2 + 1) ⋅ x = a
a2 + 1 > 0 è ïðèíèìàåò îäèíàêîâîå çíà-
x + 15 = 2x – 30
÷åíèå ïðè a > 0 è a a < 0.
45 = x;

Çíà÷èò, x= a — åäèíñòâåííûé Óïðàæíåíèÿ äëÿ ïîâòîðåíèÿ
a2 + 1
êîðåíü óðàâíåíèÿ ïðè ëþáîì a. 116. à) (x + y)2;

â) a2x = –2x; á) x2 + y2;
â) a3 – b3;
(a2 + 2) ⋅ x = 0 ã) (a – b)3;
ä) a3 + b3;
a2 + 2 ≠ 0 å) (a + b)3.

x = 0 — åäèíñòâåííûé êîðåíü óðàâíå- 117. à) ïðè a = 0,2; 5a3 = 5 ⋅ (0,2)3 =

íèÿ ïðè ëþáîì a. = 5 ⋅ 0,008 = 0,04;
á) Åñëè x = –2, òî 2x2 – x4 – 5 = 2 ⋅ 4 –
ã) 4 – 5x = a2x
a2x + 5x = 4 – 16 – 5 = –13;
â) Åñëè a = 0,2, òî a3 + 3a2 = 0,22 ⋅ (0,2 +
(a2 + 5) ⋅ x = 4
+ 3) = 0,04 ⋅ 3,2 = 0,128;
a2 + 5 ≠ 0 ã) Åñëè a = –1,2, òî 3a4 – a2 = (–1,2)2 ×
× (3 ⋅ (–1,2)2 – 1) =
x = 4 5 — åäèíñòâåííûé êîðåíü
a2 + = 1,44 ⋅ (3 ⋅ 1,44 – 1) = 1,44 ⋅ 3,32 = 4,7808;
óðàâíåíèÿ ïðè ëþáîì a. ä) Åñëè x = 2,5, y = 3, òî 1 – (x – ó)3 = 1 –

114*. à) kx = 8

x = 8 — åäèíñòâåííûé êîðåíü óðàâíåíèÿ – (2,5 – 3)3 = 1 − ⎛ 1⎞3 = 1 − 1 = 7 = 0, 875.
k ⎜⎝ 2⎟⎠ 8 8

ïðè k ≠ 0

êîðíåé íåò ïðè k = 0; 118. (10 + 11 + 12 + … + 19) + (20 +
á) (k + 3) ⋅ x = 5;
x = 5 — åäèíñòâåííûé êîðåíü + 21 + … + 29) + (30 + 31 + … + 39) +

k+3 + (40 + … + 49) + … + (90 + 91 + … + 99) =
óðàâíåíèÿ ïðè k ≠ –3;
= 10 + 19 ⋅ 10 + 20 + 29 ⋅ 10 +
êîðíåé íåò ïðè k = –3; 22

18 АЛГЕБРА (к учебнику Г. П. Бевза, В. Г. Бевз) â) k ⋅ x = k; + 30 + 39 ⋅ 10 + ... + 90 + 99 ⋅ 10 =
22
x = k = 1 — åäèíñòâåííûé êîðåíü
k = 29 ⋅ 5 + 49 ⋅ 5 + 69 ⋅ 5 + … + 189 ⋅ 5 =

óðàâíåíèÿ ïðè k ≠ 0; áåñêîíå÷íîå ìíî- = (29 + 49 + 69 + 89 + 109 + 129 + 149 +

æåñòâî êîðíåé ïðè k = 0. + 169 + 189) ⋅ 5 = (29 + (29 + 20) + (29 +

ã) (2 – k) ⋅ x = 2 – k + 40) + (29 + 60) + (29 + 80) + (29 +

x = 2 − k = 1 — ºäèíñòâåííûé êîðåíü + 100) + (29 + 120) + (20 + 140) + (29 +
2−k
+ 160)) ⋅ 5 = (29 ⋅ 9 + 720) ⋅ 5 = (261 +
ïðè k ≠ 2;
+ 720) ⋅ 5 = 4905.

áåñêîíå÷íîå ìíîæåñòâî êîðíåé ïðè k = 2. Èëè:

115*. à) 5x – 4 + 2x = 1 èìååò îäèí Ýòî àðèôìåòè÷åñêàÿ ïðîãðåññèÿ 10 + 11

+ 12 + … + 89 + 90 + … + 99, ãäå âñåãî

êîðåíü 7x = 5; x = 5 ; 90 ñëàãàåìûõ.
7
a1 = 10 10 + 99 ⋅ 90 =
5x – 4 + 2x = 7x – 4 èìååò áåñêîíå÷íîå a90 = 99, òî åñòü S90 = 2

ìíîæåñòâî êîðíåé; = 109 ⋅ 45 = 4905.

5x – 4 + 2x = 7x êîðíåé íåò. 119. à) 20 % îò 350 : (350 ⋅ 20) : 100 = 70;

á) 2 ⋅ (1,5x – 7) – 3x = x óðàâíåíèå èìå- á) 30 % îò 5600 — ýòî (5600 ⋅ 30) : 100 =
= 1680;
åò îäèí êîðåíü; â) 12 % îò 0,75 — ýòî (0,75 ⋅ 12) : 100 =
= 0,09;
2 ⋅ (1,5x – 7) – 3x = –14 èìååò áåñêîíå÷- ã) 125 % îò 1,4 — ýòî (1,4 ⋅ 1,25) : 100 =
= 1,75;
íîå ìíîæåñòâî êîðíåé; ´) 15 % îò 124 ãðí — ýòî (124 ⋅ 15) :
: 100 = 18,6;
2 ⋅ (1,5x – 7) – 3x = 0 êîðíåé íåò. ä) 48 % îò 3,5 ì — ýòî (3,5 ⋅ 48) :
: 100 = 1,68.
â) 3x + 2 = 2 óðàâíåíèå èìååò îäèí
5
êîðåíü; 3x + 2 = 3 ⋅ x = 2 ⇒ 0 ⋅ x = 0
55 5

óðàâíåíèå èìååò áåñêîíå÷íîå ìíîæåñòâî

ðåøåíèé (x — ëþáîå).