alimov-7-algebra-gdz (Алгебра - 7 класс - Алимов), страница 8

Пусть х человек учатся в школе Пифагора, из них:пребывают в молчании и 3 женщины. Составим уравнение:111⎛1 1 1 ⎞x + x + x + 3 = x ; ⎜ + + − 1⎟ ⋅ x + 3 = 0⎝2 4 7 ⎠24714 + 7 + 4 − 283⋅ x = −3 ;x = 3 4 x = 282828Ответ: 28 человек.23311. Пусть прошло х ч., осталось (12 – х) ч., это равно 2 ⋅ x .12 − x =4173 36x ; 12 = x ; x = 12 ⋅ == 5 ч.3777317Ответ: 5 ч.312. Пусть в автобусе было n чел., на первых двух остановках вышло 2m человек. Тогда после I и II остановок осталось (n − 2m) чел. Пусть на III остановке вошло х чел., тогда вавтобусе стало (n − 2m + x ) чел. = k чел.n − 2 m + x = k ; x = k − n + 2mОтвет: k − n + 2m человек.630,1 − 2 x 2 ,5 − 10 x=0,4121,2 − 24 x = 1 − 4 x20 x = 0,29 − x 2x − 3;=102313.

1)2)9 − x = 10 x − 15 ;11x = 24 ;2x=2 ;11x = 0,01()()314. 1) 12 ⋅ 52n +1 − 8 ⋅ 52 n + 4 ⋅ 52n −1 : 4 ⋅ 52 n − 2 =2 n +1− 2 n + 22n − 2n + 2− 2 ⋅5+2= 3⋅5= 5 ⋅ (75 − 10 + 1) = 5 ⋅ 66 = 3302 n −1− 2 n + 2(= 3 ⋅ 53 − 2 ⋅ 52 + 5 =)2) 36 ⋅18 n − 8 ⋅ 2 n − 4 ⋅ 9 n − 3 n +1 ⋅ 6 n +1 : 18 n −1 =⎛⎝⎞⎠1212= ⎜ 36 ⋅ 18n − ⋅ 18n − 18n +1⎟ :18n −1 = 36 ⋅ 18 − ⋅ 18 − 18n +1− n +1 =⎛⎝12= 36 ⋅ 18 − ⋅ 18 − 182 = 18 ⋅ ⎜ 36 −11⎞− 18⎟ = 18 ⋅ 17 = 315⎠22315. Т.к. 2 ⋅ (a + 1) ⋅ (b + 1) = 2 ⋅ (ab + b + a + 1) = 2ab + 2b + 2a + 2 и(a + b) ⋅ (a + b + 2) = a 2 + ab + ab + b2 + 2a + 2b == a 2 + 2ab + b2 + 2a + 2b , то из 2ab + 2b + 2a + 2 == a 2 + 2ab + b2 + 2a + 2b. , выходит, что a 2 + b2 = 2 , ч.т.д.316. 1 год – вклад а рублей, после окончания года – 1,02а руб., после окончания второго года – 1,022а.

Еще через год сумма будет равна:a ⋅ (1,02) + 0,02 ⋅ (1,02) a = a ⋅ (1,02) ⋅ (1 + 0,02) = a ⋅ (1,02) ⋅ 1,02 =222= a ⋅ (1,02) , ч.т.д.3317. n = 3:n = 5:n = 10:641000 ⋅ (1,02) = 1000 ⋅ 1,061208 ≈ 1061,2131000 ⋅ (1,02) ≈ 1000 ⋅ 110408,≈ 1104 ,151000 ⋅ (1,02)10≈ 1000 ⋅ 1,21899 ≈ 1218,992Глава IV. Разложениемногочленов на множители§ 19. Вынесение общего множителя за скобки381438⎛⎝143⎞8⎠381414318. 1) 14 ⋅ 1 − 4 ⋅ 1 = ⎜ 14 − 4 ⎟ ⋅ 1 = 10 ⋅ 1 = 12 ,52) 24 ⋅ 2 ,73 + 41 ⋅ 2 ,73 = (24 + 41) ⋅ 2 ,73 = 65 ⋅ 2 ,73 = 177 ,45319.

1) 2m + 2n = 2 ⋅ (m + n) ;2) 3a − 3x = 3 ⋅ (a − x )3) 8 − 4 x = 4 ⋅ (2 − x ) ;4) 6a + 12 = 6 ⋅ (a + 2)320. 1) 9a + 12b + 6 = 3 ⋅ (3a + 4b + 2)2) 21a − 7b + 42 = 7 ⋅ (3a − b + 6)3) −10 x + 15 y − 75z = 5 ⋅ ( −2 x + 3 y − 15z )4) 9 x − 3 y + 15 z = 3 ⋅ (3 x − y + 5 z )321. 1) ax − ay = a ⋅ ( x − y ) ;2) cd + bc = c ⋅ (d + b)3) xy + x = x ⋅ ( y + 1) ;4) x − xy = x ⋅ (1 − y )322.

1) 9mn + 9n = 9n ⋅ (m + 1) ;2) 3bd − 3b = 3b ⋅ (d − 1)3) 11z − 33 yz = 11z ⋅ (1 − 3 y ) ;()4) 6 pk − 3 p = 3 p ⋅ (2 k − 1)2) a 4 − 3a 3 = a 3 ⋅ (a − 3)323. 1) a 4 + 2a 2 = a 2 ⋅ a 2 + 2 ;()4) x 2 y 3 − x 3 y 2 = x 2 y 2 ⋅ ( y − x )3) a 4b2 + ab3 = ab2 ⋅ a 3 + b ;324. 1) 9a 2b2 − 12ab3 = 3ab2 ⋅ (3a − 4b) ;2) 20 x 3 y 2 + 4 x 2 y = 4 x 2 y ⋅ (5xy + 1)()+ 3xy )325. 1) 4a 2b2 + 36a 2b3 + 6ab4 = 2ab2 ⋅ 2a + 18ab + 3b22 44 23 32 2(22) 2 x y − 2 x y + 6 x y = 2 x y ⋅ y − x2326. 1) ab − ac + a 2 = a ⋅ (b − c + a ) ;2) xy − x 2 + xz = x ⋅ ( y − x + z )653) 6a 2 − 3a + 12ba = 3a ⋅ (2a − 1 + 4b)4) 4b 2 + 8ab − 12a 2 b = 4b ⋅ (b + 2a − 3a 2 )327.

1) 1372 + 137 ⋅ 63 = 137 ⋅ (137 + 63) = 137 ⋅ 200 = 274002) 1872 − 187 ⋅ 87 = 187 ⋅ (187 − 87) = 187 ⋅ 100 = 187003) 0,7 3 + 0,7 ⋅ 9,51 = 0,7 ⋅ (0,49 + 9,51) = 0,7 ⋅10 = 74) 0,93 − 0,81 ⋅ 2,9 = 0,81 ⋅ (0,9 − 2,9) = 0,81 ⋅ (−2) = −1,62328. 1) a ⋅ (m + n ) + b ⋅ (m + n ) = (m + n ) ⋅ (a + b )2) b ⋅ (a + 5) − c ⋅ (a + 5) = (a + 5) ⋅ (b − c)3) a ⋅ (b − 5) − (b − 5) = (b − 5) ⋅ (a − 1)4) ( y − 3) + b ⋅ ( y − 3) = ( y − 3) ⋅ (1 + b)329. 1) 2a ⋅ (a − b) + 3b ⋅ (a − b) = (a − b) ⋅ (2a + 3b)2) 3n ⋅ (m − 3) + 5m ⋅ (m − 3) = (m − 3) ⋅ (3n + 5m)3) 5a ⋅ ( x + y ) − 4b ⋅ ( x + y ) = ( x + y ) ⋅ (5a − 4b)4) 7a ⋅ (c − d ) − 2b ⋅ (c − d ) = (c − d ) ⋅ (7a − 2b)()2) a ⋅ (x + y ) + b ⋅ (x + y ) = (x + y ) ⋅ (a + b )3) a ⋅ ( x + y ) − b ⋅ ( x + y ) = ( x + y ) ⋅ (a − b)4) x ⋅ ( a + 2b ) + y ⋅ ( a + 2b ) = ( a + 2b ) ⋅ ( x + y )330.

1) a 2 ⋅ ( x − y ) + b2 ⋅ ( x − y ) = ( x − y ) ⋅ a 2 + b22322222222223222331. 1) c ⋅ (a − b) + b ⋅ (b − a ) = (a − b) ⋅ (c − b)2) a ⋅ (b − c) − c ⋅ (c − b) = (b − c) ⋅ (a + c)3) ( x − y ) + b ⋅ ( y − x ) = ( x − y ) ⋅ (1 − b)4) 2b ⋅ ( x − y ) − ( y − x ) = ( x − y ) ⋅ (2b + 1)332. 1) 7 ⋅ ( y − 3) − a ⋅ (3 − y ) = ( y − 3) ⋅ (7 + a )2) 6 ⋅ (a − 2) + a ⋅ (2 − a ) = (a − 2) ⋅ (6 − a )(3) b2 ⋅ (a − 1) − c ⋅ (1 − a ) = (a − 1) ⋅ b2 + c()4) a ⋅ (m − 2) + b ⋅ (2 − m ) = (m − 2) ⋅ a − b2662)333. 1) a ⋅ (b − c ) + d ⋅ (b − c ) − 7 ⋅ (c − b ) = (b − c ) ⋅ (a + d + 7 )2) x ⋅ ( x − y ) + y ⋅ ( y − x ) − 3 ⋅ ( x − y ) = ( x − y ) ⋅ ( x − y − 3)3) x ⋅ (a − 2) + y ⋅ (2 − a ) + (2 − a ) = (a − 2) ⋅ ( x − y − 1)4) a ⋅ (b − 3) + (3 − b ) − b ⋅ (3 − b ) = (b − 3) ⋅ (a − 1 + b )334. 1) 7 ⋅ (a − 5) − b ⋅ (5 − a ) = (a − 5) ⋅ (7 + b)(2 − 5) ⋅ (7 + 3) = −3 ⋅ 10 = −3022) a ⋅ (a − b) + b ⋅ (b − a ) = (a − b) ⋅ (a − b) = (a − b)a = 6,3; b = 2 ,3:(6,3 − 2,3)2 = 163) 2 x ⋅ ( x + y ) − 3 y ⋅ ( x + y ) + 7 ⋅ ( x + y ) = ( x + y ) ⋅ (2 x − 3 y + 7)a = 2; b = 3:x = 4; y = 5:(4 + 5)⋅ (2 ⋅ 4 − 3 ⋅ 5 + 7 ) = 9 ⋅ (8 − 15 + 7 ) = 04) x ⋅ ( y − x ) − y ⋅ ( x − y ) − 4 ⋅ ( y − x ) = ( y − x ) ⋅ ( x + y − 4)x = 3; y = −5:(−5 − 3) ⋅ (3 − 5 − 4) = 8 ⋅ 6 = 48335.

1) 3( x + y ) ⋅ ( x − y ) − ( x + y ) = ( x + y ) ⋅ (3x − 3 y − x − y ) =2= ( x + y ) ⋅ (2 x − 4 y ) = 2 ⋅ ( x + y ) ⋅ ( x − 2 y )2) 5 ⋅ (a − b) − (a + b) ⋅ (b − a ) = (a − b) ⋅ (5a − 5b + a + b) =2= (a − b) ⋅ (6a − 4b) = 2 ⋅ (a − b) ⋅ (3a − 2b)3) ( x + y ) − x ⋅ ( x + y ) = ( x + y ) ⋅ ( x + y − x ) = y ⋅ ( x + y )32224) a ⋅ (a − b) − (b − a ) = (a − b) ⋅ ( −b + a ) = (a − b) ⋅ (2a − b)2322336. 1) x 2 ⋅ ( x − 3) − x ⋅ ( x − 3) = x ⋅ ( x − 3) ⋅ ( x − x + 3) = 3x ⋅ ( x − 3)22) a 3 ⋅ (2 + a ) + a 2 ⋅ (2 + a ) = a 2 ⋅ (2 + a ) ⋅ ((a + 2) + a ) =2= a 2 ⋅ (2 + a ) ⋅ (2a + 2) = 2a 2 ⋅ (2 + a ) ⋅ (a + 1)3) 3m ⋅ (n − m) − 9m2 ⋅ (m − n) = 3m ⋅ (m − n) ⋅ (m − n − 3m) =2= 3m ⋅ (m − n) ⋅ ( − n − 2m) = 3m ⋅ (n − m) ⋅ (n + 2m)4) 15 p2 ⋅ ( p + q ) − 5 p ⋅ ( p + q ) = 5 p ⋅ ( p + q ) ⋅ (3 p − p − q ) =2= 5 p ⋅ ( p + q ) ⋅ (2 p − q )67337.

1) x 2 − 2 x = 0 ;x ⋅ ( x − 2) = 0 ;2) 3x + x 2 = 0x ⋅ (3 + x ) = 0x −2 = 0;x1 = 2 ; x2 = 0 ;3+ x = 0x1 = −3 ; x 2 = 03) 5x 2 + 3x = 0 ;x ⋅ (5x + 3) = 0 ;4) 4 x 2 − 7 x = 05x + 3 = 0 ;3x1 = − ; x2 = 0 ;54x − 7 = 07x1 = ; x2 = 04x ⋅ (4 x − 7) = 05) x 2 ⋅ ( x − 2) − 2 x ⋅ ( x − 2) = 0 ; 6) 3x ⋅ (1 − x ) − x 2 ⋅ (1 − x ) = 022x ⋅ ( x − 2 ) ⋅ ( x − 2 x + 4) = 0 ;x ⋅ (1 − x ) ⋅ (3 − 3x − x ) = 0−x + 4 = 0 ;x ⋅ (1 − x ) ⋅ (3 − 4 x ) = 0x1 = 4 ; x2 = 2 ; x3 = 0 ;x1 =3; x2 = 1 ; x3 = 04338.

Пусть х – данное число; т.к. x M на 225 и в остатке получается150, то x = 225a + 150 = 75 ⋅ (3a + 2)M75 , т.к.75⋅ (3a + 2) : 75 = 3a + 2 , ч.т.д.§ 20. Способ группировки339. 1) a + b + c ⋅ (a + b) = (a + b) ⋅ (1 + c)2) m − n + p ⋅ (m − n) = (m − n) ⋅ (1 + p)3) x + 3a ⋅ ( x + y ) + y = ( x + y ) ⋅ (1 + 3a )4) x + 2a ⋅ ( x − y ) − y = ( x − y ) ⋅ (1 + 2a )340. 1) 2m ⋅ (m − n) + m − n = (m − n) ⋅ (2m + 1)2) 4q ⋅ ( p − 1) + p − 1 = ( p − 1) ⋅ (4q + 1)3) 2m ⋅ (m − n) + n − m = (m − n) ⋅ (2m − 1)4) 4q ⋅ ( p − 1) + 1 − p = ( p − 1) ⋅ (4q − 1)341.

1) ac + bc − 2ad − 2bd = c ⋅ (a + b) − 2d ⋅ (a + b) = (a + b) ⋅ (c − 2d )2) ac − 3bd + ad − 3bc = a ⋅ (c + d ) − 3b ⋅ (c + d ) = (c + d ) ⋅ (a − 3b)683) 2bx − 3ay − 6by + ax = 2b ⋅ ( x − 3 y ) + a ⋅ ( x − 3 y ) = ( x − 3 y ) ⋅ (2b + a )4) 5ay − 3bx + ax − 15by = a ⋅ (5 y + x ) − 3b ⋅ ( x + 5 y ) = (5 y + x ) ⋅ (a − 3b)342. 1) 18a 2 − 27ab + 14ac − 21bc = 9a ⋅ (2a − 3b) + 7c ⋅ (2a − 3b) == (2a − 3b) ⋅ (9a + 7c)()2) 10 x 2 + 10 xy + 5 x + 5 y = 10 x 2 + 10 xy + (5 x + 5 y ) == 10 ⋅ ( x + y ) + 5 ⋅ ( x + y ) = 5 ⋅ ( x + y ) ⋅ (2 x + 1)3) 35ax + 24 хy − 20ay − 42 x 2 = 7 x ⋅ (5a − 6 х ) + 4 y ⋅ (6 x − 5а ) == (5a − 6 x ) ⋅ (7 x − 4 y )()4) 48 xz 2 + 32 xy 2 − 15 yz 2 − 10 y 3 = 16 x ⋅ 3 z 2 + 2 y 2 −() ()− 5 y ⋅ 3z 2 + 2 y 2 = 3 z 2 + 2 y 2 ⋅ (16 x − 5 y )() ()= 16a ⋅ ( b + 2c ) − 5c ⋅ ( b + 2c ) = ( b + 2c ) ⋅ (16a − 5c)Проверим: ( b + 2c ) ⋅ (16a − 5c) = 16ab + 32ac − 5b c − 10c343.

1) 16ab 2 − 5b 2 c − 10c 3 + 32ac 2 = 16ab 2 + 32ac 2 − 5b 2 c + 10c 3 =2222222222232) 6mnk 2 + 15m 2 k − 14n 3 k − 35mn 2 = 3mk ⋅ (2nk + 5m ) −(− 7n 2 ⋅ (2nk + 5m ) = (2nk + 5m ) ⋅ 3mk − 7n 2(Проверим: (2nk + 5m) ⋅ 3mk − 7n2)) = 6nk m + 15m k − 14n k − 35nm2233) −28ac + 35c2 − 10cx + 8ax = 7c ⋅ (5c − 4a ) + 2 x ⋅ (4a − 5c) == (5c − 4a ) ⋅ (7c − 2 x )Проверим: (5c − 4a ) ⋅ (7c − 2 x ) = 35c 2 − 28ac − 10cx + 8ax4) −24bx − 15c2 + 40bc + 9cx = 8b ⋅ (5c − 3x ) + 3c ⋅ (3x − 5c) == (5c − 3x ) ⋅ (8b − 3c)Проверим: (5c − 3x ) ⋅ (8b − 3c) = 40bc − 24bx − 15c2 + 9cx() () ()344.

1) xy 2 − by 2 − ax + ab + y 2 − a = x ⋅ y 2 − a − b ⋅ y 2 − a + y 2 − a =()= y − a ⋅ ( x − b + 1)2()Проверим: y 2 − a ⋅ ( x − b + 1) = xy 2 − ax − by 2 + ba + y 2 − a692) ax 2 − ay − bx 2 + cy + by − cx 2 = (a − b − c) ⋅ x 2 − (a − b − c) ⋅ y == (a − b − c ) ⋅ ( x 2 − y )()Проверим: (a − b − c) ⋅ x 2 − y = ax 2 − bx 2 − cx 2 − ay + by + cy2 22223) a x − bx + a x − bx + a y − by == a 2( x 2 + x + y ) − b( x 2 + x + y ) = (a 2 − b)( x 2 + x + y )Проверим: (a 2 − b)( x 2 + x + y ) = a 2x 2 − bx 2 + a 2x − bx + a 2 y − by4) ax 2 − bx 2 + ay − by − ax + bx = a( x 2 + y − x) − b( x 2 + y − x) == (a − b)( x 2 + y − x)Проверим: (a − b)( x 2 + y − x) = ax 2 − bx 2 + ay − by − ax + bx345. 1) 5a 2 − 5ax − 7a + 7 x = 5a ⋅ (a − x ) − 7 ⋅ (a − x ) = (a − x ) ⋅ (5a − 7 )(4 + 3) ⋅ (5 ⋅ 4 − 7) = 7 ⋅ 13 = 912) m − mn − 3m + 3n = m ⋅ (m − n) − 3 ⋅ (m − n) = (m − n) ⋅ (m − 3)m = 0,5; n = 0,25:(0,5 − 0,25) ⋅ (0,5 − 3) = 0,25 ⋅ ( −2 ,5) = −0,6253) a 2 + ab − 5a − 5b = a ⋅ (a + b) − 5 ⋅ (a + b) = (a + b) ⋅ (a − 5)a = 6,6; b = 0,4 :(6,6 + 0,4) ⋅ (6,6 − 5) = 7 ⋅ 1,6 = 11,224) a − ab − 2a + 2b = a ⋅ (a − b) − 2 ⋅ (a − b) = (a − b) ⋅ (a − 2)x = −3; a = 4 :2a=77 1572013; b = 0,15: ( −) ⋅ ( − 2) =⋅ (−1 ) =2020 100 2010020=−33= −0,33100346.

1) 139 ⋅ 15 + 18 ⋅ 139 + 15 ⋅ 261 + 18 ⋅ 261 == 139 ⋅ (15 + 18) + 261 ⋅ (15 + 18) = 33 ⋅ (139 + 261) = 33 ⋅ 400 = 132002) 125 ⋅ 48 − 31 ⋅ 82 − 31 ⋅ 43 + 125 ⋅ 83 = 125 ⋅ (48 + 83) − 31 ⋅ (82 + 43) == 125 ⋅ 131 − 31 ⋅ 125 = 125 ⋅ (131 − 31) = 125003) 14,7 ⋅ 13 − 2 ⋅ 14 ,7 + 13 ⋅ 5,3 − 2 ⋅ 5,3 == 14 ,7 ⋅ (13 − 2) + 5,3 ⋅ (13 − 2) = 11 ⋅ (14 ,7 + 5,3) = 11 ⋅ 20 = 2201 121 421144) 3 ⋅ 4 + 4,2 ⋅ + 3 ⋅ 2 + 2,8 ⋅ = 3 ⋅ (4 + 2 ) +3 533 533552121 2+ ⋅ (4,2 + 2,8) = 3 ⋅ 7 + ⋅ 7 = 7 ⋅ (3 + ) = 283333 370((2) x 2 + 7 х − 4 x − 28 = 0x ⋅ ( x − 4) + ( x − 4) = 0 ;x ⋅ ( x + 7) − 4 ⋅ ( x + 7) = 0( x − 4) ⋅ ( x + 1) = 0 ;( x + 7) ⋅ ( x − 4 ) = 0x + 1 = 0; x − 4 = 0 ;x − 4 = 0;x1 = −1; ;x1 = 4;x2 = 4x 2 = −73) 5x 2 − 10 x + ( x − 2) = 0 ;348.))347.