Bài 6: Phân tích đa thức thành nhân tử bằng phương pháp đặt nhân tử chung

HM
23 tháng 8 2023 lúc 17:58

Bài 1A:

\(a,x^2-xy+x-y\\ =x\left(x-y\right)+\left(x-y\right)\\ =\left(x+1\right)\left(x-y\right)\\ b,xz+yz+4x+4y\\ =z\left(x+y\right)+4\left(x+y\right)\\ =\left(x+y\right)\left(z+4\right)\\ c,x^2-x-y^2+y\\ =\left(x-y\right)\left(x+y\right)-\left(x-y\right)\\ =\left(x-y\right)\left(x+y-1\right)\\ d,x^2+2x+2z-z^2\\ =\left(x-z\right)\left(x+z\right)+2\left(x+z\right)\\ =\left(x+z\right)\left(x-z+2\right)\)

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HM
23 tháng 8 2023 lúc 18:00

Bài 1B:

\(a,x^2+2xy+x+2y\\ =x\left(x+2y\right)+\left(x+2y\right)\\ =\left(x+2y\right)\left(x+1\right)\\ b,2xy+yz+2x+z\\ =y\left(2x+z\right)+\left(2x+z\right)\\ =\left(y+1\right)\left(2x+z\right)\\ c,y^2-2y-z^2-2z\\ =\left(y-z\right)\left(y+z\right)-2\left(y+z\right)\\ =\left(y+z\right)\left(y-z-2\right)\\ d,x^3-x-y+y^3\\ =\left(x+y\right)\left(x^2-xy+y^2\right)-\left(x+y\right)\\ =\left(x+y\right)\left(x^2-xy+y^2-1\right)\)

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H24
23 tháng 8 2023 lúc 18:01

1B

a)

\(=x\left(x+2y\right)+\left(x+2y\right)\\ =\left(x+1\right)\left(x+2y\right)\)

b)

\(=y\left(2x+z\right)+\left(2x+z\right)\\ =\left(y+1\right)\left(2x+z\right)\)

c)

\(=\left(y^2-z^2\right)-\left(2y+2z\right)\\ =\left(y-z\right)\left(y+z\right)-2\left(y+z\right)\\ =\left(y+z\right)\left(y-z-2\right)\)

d)

\(=\left(x^3+y^3\right)-\left(x+y\right)\\ =\left(x+y\right)\left(x^2-xy+y^2\right)-\left(x+y\right)\\ =\left(x+y\right)\left(x^2-xy+y^2-1\right)\)

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