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a) x2-2x-y2+2y
=(x2-y2)-(2x-2y)
=(x-y)(x+y)-2(x-y)
=(x-y)(x+y-2)
![](https://rs.olm.vn/images/avt/0.png?1311)
a) \(3x^2-6xy+3y^2-12x^2=3\left(x^2-2xy+y^2\right)-12x^2=3\left(x-y\right)^2-12x^2=3\left[\left(x-y\right)^2-4x^2\right]=3\left(x-y-2x\right)\left(x-y+2x\right)=3\left(-x-y\right)\left(3x-y\right)\)
b)\(3x^2y^2-6x^2y^3+12x^2y^2=3x^2y^2\left(1-2y+4\right)=3x^2y^2\left(5-2y\right)\)
c) \(3x^2-3y^2+12x-12y=3\left(x^2-y^2\right)+12\left(x-y\right)=3\left(x-y\right)\left(x+y+4\right)\)
a: \(3x^2-6xy+3y^2-12x^2\)
\(=3\left(x^2-2xy+y^2-4x^2\right)\)
\(=3\left[\left(x-y\right)^2-4x^2\right]\)
\(=3\left(x-y-2x\right)\left(x-y+2x\right)\)
\(=3\left(-x-y\right)\left(3x-y\right)\)
b: \(3x^2y^2-6x^2y^3+12x^2y^2\)
\(=3x^2y^2\left(1-2y+4\right)\)
\(=3x^2y^2\left(-2y+5\right)\)
c: Ta có: \(3x^2-3y^2+12x-12y\)
\(=3\left(x-y\right)\left(x+y\right)+12\left(x-y\right)\)
\(=3\left(x-y\right)\left(x+y+4\right)\)
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3x2 + 6xy + 3y2 – 3z2
= 3.(x2 + 2xy + y2 – z2)
(Nhận thấy xuất hiện x2 + 2xy + y2 là hằng đẳng thức nên ta nhóm với nhau)
= 3[(x2 + 2xy + y2) – z2]
= 3[(x + y)2 – z2]
= 3(x + y – z)(x + y + z)
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b, (\(x^2\) - \(xy\) ) + (\(x-y\))
= (\(x-y\)).\(x\) + (\(x-y\))
= (\(x-y\)).(\(x\) + 1)
c, \(x^2\) - 2\(x\) - 15
= (\(x^2\) - 2\(x\) + 1) - 16
= (\(x\) - 1)2 - 42
= (\(x-1-4\)).(\(x-1+4\))
= (\(x-5\)).(\(x+3\))
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a/ x2 + 4x - 21= x2 - 3x +4x - 21
= (x2+4x)-(3x+21)
= x(x+4)- 3(x+7)
= (x-3).(x+7)
b/ 3x2-6xy+3y2-3z2 = 3(x2- 2xy+y2- z2)
= 3[(x2 + 2xy + y2) – z2]
= 3[(x + y)2 – z2]
= 3(x + y – z)(x + y + z)
c/ 2x2y + 12xy + 18y = 2y(x2+6x+9)
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`@` `\text {Ans}`
`\downarrow`
`a,`
`3x^2 + 6xy + 3y^2 - 3z`
`= 3*x^2 + 3*2xy + 3y^2 - 3z`
`= 3(x^2 + 2xy + y^2 - z)`
`b,`
`x^3 + x^2y - x^2z - xyz`
`= x(x + y)(x-z)`
![](https://rs.olm.vn/images/avt/0.png?1311)
\(3x^4y-12x^2y^3=3x^2y\left(x^2-4y^2\right)=3x^2y\left(x-2y\right)\left(x+2y\right)\)
\(x^2-y^2-8y-16=x^2-\left(y^2+8y+16\right)=x^2-\left(y+4\right)^2=\left(x+y+4\right)\left(x-y-4\right)\)
\(x^3+3x^2+4x+12=x^2\left(x+3\right)+4\left(x+3\right)=\left(x^2+4\right)\left(x+3\right)\)
\(3x^2-6xy+3y^2-27=3\left[\left(x-y\right)^2-9\right]=3\left(x-y-3\right)\left(x-y+3\right)\)
\(3x^2-6xy+3y^2-12z^2\)
\(=3\left(x^2-2xy+y^2-4z^2\right)\)
\(=3\left[\left(x-y\right)^2-4z^2\right]=3\left(x-y-2z\right)\left(x-y+2z\right)\)
Ta có: \(3x^2-6xy+3y^2-12z^2\)
\(=3.\left(x^2-2xy+y^2-4z^2\right)\)
\(=3.\left[\left(x-y\right)^2-4z^2\right]\)
\(=3.\left(x-y-2z\right).\left(x-y+2z\right)\)