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
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
`\text{~~Holi~~}`
`a. x(x+5)-x-5`
`= (x+5)(x-1)`
`b. x^2+1/6x-1/6`
`= 1/6 .(6x^2+x-1)`
`= 1/6 .(6x^2+3x-2x-1)`
`= 1/6 .[3x(2x+1)-(2x+1)]`
`= 1/6 .(2x+1)(3x-1)`
`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)`
`f. x^3-10x^2+25`
`= x^3-5x^2-5x^2+25`
`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[(x-1)^2-y^2]`
`= x(x-1-y)(x-1+y)`
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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