(1)-2ab2+4a2b-10b (2)x2y2+y4+x4 (3)25a4-10a2+1 (4)x2-x-6 (5) xy-xz+y-z (6)16-(x+y)2

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(1)-2ab2+4a2b-10b (2)x2y2+y4+x4 (3)25a4-10a2+1 (4)x2-x-6 (5) xy-xz+y-z (6)16-(x+y)2
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(1)-2ab2+4a2b-10b (2)x2y2+y4+x4 (3)25a4-10a2+1 (4)x2-x-6 (5) xy-xz+y-z (6)16-(x+y)2
(1)-2ab2+4a2b-10b (2)x2y2+y4+x4 (3)25a4-10a2+1 (4)x2-x-6 (5) xy-xz+y-z (6)16-(x+y)2

(1)-2ab2+4a2b-10b (2)x2y2+y4+x4 (3)25a4-10a2+1 (4)x2-x-6 (5) xy-xz+y-z (6)16-(x+y)2
(1)-2ab²+4a²b-10b
= -2b(b²-2b+5)
(2)x²y²+y4+x4
=(x²+y²)² - x²y²
=(x²+xy+y²)(x²-xy+y²)
(3)25a4-10a2+1
=(5a² - 1)²
=(√5a+1)²(√5a-1)²
(4)x2-x-6
=(x-3)(x+2)
(5) xy-xz+y-z
=(x+1)(y-z)
(6)16-(x+y)2
=(4-x-y)(4+x+y)

我建议您把问题重写一遍,几次方用^x表示

(1).所有之前都为-2的倍数,所以提取-2,在提取b
以下类同

(1)原式=2b(2a^2-ab-5b);(2)原式=x^4+2x^2y^2+y^4-x^2y^2=(x^2+y^2)^2-(xy)^2=(x^2-xy+y^2)(x^2+xy+y^2);
(3)原式=(5a^2-1)^2=(根号倍a+1)^2*(根号倍a-1)^2;(4)原式=(x-3)(x+2);(5)原式=(xy-xz)+(y-z)=x(x-z)+(y-z)=(x-z)(x+1);(6)原式=4^2-(x+y)^2=(4+x+y)(4-x-y).

(1)-2ab2+4a2b-10b=2b(-ab+2a²-5)=2b(2a²-ab-5)
(2)x2y2+y4+x4= 2x2y2+y4+x4-x2y2=(x²+y²)²-(xy)²=(x²+xy+y²)(x²-xy+y²)
(3)25a4-10a2+1 =(5a²-1)&...

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(1)-2ab2+4a2b-10b=2b(-ab+2a²-5)=2b(2a²-ab-5)
(2)x2y2+y4+x4= 2x2y2+y4+x4-x2y2=(x²+y²)²-(xy)²=(x²+xy+y²)(x²-xy+y²)
(3)25a4-10a2+1 =(5a²-1)²=(√5x+1)²(√5x-1)²------注:最后括号里面的根式里是5,x在根式外。
(4)x2-x-6 =(x-3)(x+2)
(5) xy-xz+y-z =(xy-xz)+(y-z)=x(y-z)+(y-z)=(x+1)(y-z)
(6)16-(x+y)2=4²-(x+y)²=(4+x+y)(4-x-y)

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