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\(3x^4y-12x^2y^3=3x^2y\left(x^2-4y^2\right)=3x^2y\left(x-2y\right)\left(x+2y\right)\)
\(x^2-y^2-8y-16=x^2-\left(y^2+8y+16\right)=x^2-\left(y+4\right)^2=\left(x+y+4\right)\left(x-y-4\right)\)
\(x^3+3x^2+4x+12=x^2\left(x+3\right)+4\left(x+3\right)=\left(x^2+4\right)\left(x+3\right)\)
\(3x^2-6xy+3y^2-27=3\left[\left(x-y\right)^2-9\right]=3\left(x-y-3\right)\left(x-y+3\right)\)
a)\(A=3x^2+6xy+3y^2-3z^2=3\left(x^2+2xy+y^2-z^2\right)=3\left[\left(x+y\right)^2-z^2\right]=3\left(x+y-z\right)\left(x+y+z\right)\)b) \(A=\left(x+y\right)^2-2\left(x+y\right)+1=\left(x+y-1\right)^2\)
c) \(A=x^2+y^2+2xy+yz+zx=\left(x+y\right)^2+z\left(x+y\right)=\left(x+y\right)\left(x+y+z\right)\)
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: \(3x^2-3y^2+12x-12y\)
\(=3\left(x-y\right)\left(x+y\right)+12\left(x-y\right)\)
\(=3\left(x-y\right)\left(x+y+4\right)\)
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)
`@` `\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)`
\(3x^2+6xy+3y^2=3\cdot\left(x^2+2xy+y^2\right)=3\cdot\left(x+y\right)^2\)
a) x2-2x-y2+2y
=(x2-y2)-(2x-2y)
=(x-y)(x+y)-2(x-y)
=(x-y)(x+y-2)
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)
a, ( x-y)2=4
3x^2 +3y^2 -6xy -12
=3(x^2 - 2xy +y^2 - 2^2 )
=3 (x-y)^2 - 2^2
=3(x-y-2)(x-y+2)
3(x+y) -(x^2+2xy+y^2)
=3(x+y) -(x+y)^2
(x+y)(3-x-y)