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Ta có : \(x^3+2x^2y+xy^2-9x\)
\(=x\left(x^2+2xy+y^2-9\right)\)
\(=x\left[\left(x^2+2xy+y^2\right)-9\right]\)
\(=x\left[\left(x+y^2\right)-3^2\right]\)
\(=x\left(x+y-3\right)\left(x+y+3\right)\)
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3 . (x+3) - x2+9
= 3.(x+3)-(x2-9)
=3.(x+3)-[(x-3).(x+3)]
=(x+3).[3-(x-3)]
=(x+3).(3-x+3)
=(x+3).(9-x)
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Ta có:\(9\left(x+5\right)^2-4\left(x-7\right)^2=9\left(x^2+10x+25\right)-4\left(x^2-14x+49\right)\)
\(=9x^2+90x+225-4x^2+56x-196=5x^2+146x+29\)
\(=\left(5x^2+145x\right)+\left(x+29\right)=\left(x+29\right)\left(5x+1\right)\)
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Bài làm:
Ta có: \(9\left(x-y\right)^2-4\left(x+y\right)^2=\left[3\left(x-y\right)\right]^2-\left[2\left(x+y\right)\right]^2\)
\(=\left(3x-3y-2x-2y\right)\left(3x-3y+2x+2y\right)=\left(x-5y\right)\left(5x-y\right)\)
Học tốt!!!!
Ta có :
\(9\left(x-y\right)^2-4\left(x+y\right)^2=9x^2-18xy+9y^2-4x^2-8xy-4y^2\)
\(=5x^2-26xy+5y^2==\left(5x-y\right)\left(x-5y\right)\)
![](https://rs.olm.vn/images/avt/0.png?1311)
a) x3-2x2-x+2
=x(x2-1)+2(-x2+1)
=x(x2-1)-2(x2-1)
=(x2-1)(x-2)
b)
x2+6x-y2+9
=x2+6x+9-y2
=(x+3)2-y2
=(x+3-y)(x+3+y)
x^2-9=x^2-3^2
=(x-3).(x+3)
`x^2-9=x^2 - 3^2=(x-3)(x+3)`