Практические задания Вынесение общего множителя за скобки» 7 класс. Вынесение общего множителя за скобки
1 вариант
1 тип. 1) 8х + 8у = 8(х + у) 2) 5х – xz = x(5 – z) 3) 4n – 4 = 4n - 4∙1 = 4(n – 1)
15х + 15у = bc + bx = 6y – 6 =
21к – 21t = 10xy – ym = 7z + 7 – 7x =
0,4a – 0,4b + 0,4n =
2 тип. 1) – 8x – 8y = 8(- x – y) 2) – 8x – 8y = - 8(x +y) [ – 8x – 8y = - 8x + (- 8y) ]
-11d - 11c = -11d - 11c =
- xa – xb = - xa – xb =
3 тип. 1) 15х + 25у = 5(3х + 5у), [ НОД(15, 25) = 5 ]
2) 3c + 21d – 30x = 3(c + 7d -10x)
5x + 5y = 9x – 12t =
12y - 18z + 30x = 8b - 8k – 24x =
4 тип. 1) 12xy - 18yxz +9yz = 3y(4x – 6xz + 3z)
2) 12xy – 7yxz + 9yz = y(12x – 7xz + 9z)
6xy + 10xz = 22bc – 11c =
8nd + 16nk – 8mn = 15nk – 20xk + 10xky =
14xz – 5yz – 11mz =
5 тип. 1) 2x(x + 2) + 5(x + 2) = (x + 2)(2x + 5)
2) 2x(x + 2) + 5(2 + x) = 2x(x + 2) + 5(x + 2) = (x + 2)(2x + 5)
3) 2x(x - 2) + 5(x – 2) = (x – 2)(2x + 5)
4) 2x(x - 2) + 5(2 - x) = 2x(x - 2) - 5(x - 2) = (x – 2)(2x - 5) [ 5 – 2 = 3; 2 – 5 = - 3 ]
5y(2x + 4) + x(2x + 4) = 13(y – 8) – z(y – 8) =
4y(2x + 1) + x(1 + 2x) = z(2y – 5) – 12(5 – 2y) =
6 тип. a5 + a3 + a2 = a∙a∙a∙a∙a + a∙a∙a + a∙a∙1 = a2 (a3 + a + 1)
x7 + x3 – x4 = c3 + c5 – c + c4 =
7 тип. 1) 4y4 – 8y2 + 6y = 2y(2y3 – 4y + 3)
2) 5x4y2 + 20x3y5 – 25x6y2 = 5x3y2 (3x + 4y3 – 5x3)
3) 12ab4 – 18a2b3c = 6ab3(2b – 3ac)
15a4b2 + 6a2b3 = 9y4 + 21y5 – 30y7 =
- 20xy2 – 24x2y + 45x2y2 =
8 тип. 1) (x – 2)2 + (x – 2) = (x – 2)(x – 2) + (x – 2)∙1 = (x – 2)(x - 2 + 1) = (x – 2)(2x - 1) =
2) (x – 2)2 + x(x – 2) = (x – 2)(x – 2) + x(x – 2) = (x – 2)(x - 2 + x) = (x – 2)(2x - 2∙1) =
= (x – 2)∙2∙(x – 1) = 2(x – 2)(x – 1)
3) (2 - x)2 + x(x – 2) = (x - 2)2 + x(x – 2) = (x – 2)(x - 2 + x) = (x – 2)(2x - 2∙1) =
= (x – 2)∙2∙(x – 1) = 2(x – 2)(x – 1)
4) 14(x – 2)2 + x(x – 2) = (x – 2)∙(14(x – 2) + x) = (x – 2)(14x – 28 + x) = (x – 2)(15x – 28)
(2m + 3) + 5(2m + 3)2 =
2y(y +3) + 6(3+ y)2 =
-3(2x +1)2 - (2x + 1) =
5(4 – z)2 – 4(4 – z) =
y(2y – 7) + 2(7 – 2y)2 =
