Phân tích đa thức thành nhân tử: a)x(x+5)-x-5 b)x²+1/6x-1/6 c)x³-2x²+x d)x²-y²-4x+4y e)3x²-3xy_7x+7y f)x³-10x²+25 g)x³-10x²+25x h)x³-2x²+x+xy²

Đáp án:

a) $x(x+5)-x-5=x(x+5)-(x+5)=(x+5)(x-1)$

b) $x^2+\dfrac{1}{6}x-\dfrac{1}{6}=x^2+2.\dfrac{1}{12}+\left(\dfrac{1}{12}\right)^2-\dfrac{25}{144}=\left(x+\dfrac{1}{12}\right)^2-\left(\dfrac{5}{12}\right)^2=\left(x+\dfrac{1}{12}+\dfrac{5}{12}\right)\left(x+\dfrac{1}{12}-\dfrac{5}{12}\right)=\left(x+\dfrac{1}{2}\right)\left(x-\dfrac{1}{3}\right)$

c) $x^3-2x^2+x=x(x^2-2x+1)=x(x-1)^2$

d) $x^2-y^2-4x+4y=(x-y)(x+y)-4(x-y)=(x-y)(x+y-4)$

e) $3x^2-3xy-7x+7y=3x(x-y)-7(x-y)=(x-y)(3x-7)$

g) $x^3-10x^2+25x=x(x^2-10x+25)=x(x-5)^2$

h) $x^3-2x^2+x-xy^2=x(x^2-2x+1-y^2)=x\left[(x-1)^2-y^2\right]=x(x-1-y)(x-1+y)$

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