Đề si thì phải 12x^2 - 12xy + 3y^2 

sai de

3*(\(4x^2-4xy+y^2\))-10(2x-y)+8

3*(2x-y)^2-10(2x-y)+8

3*(2x-y)^2-6(2x-y)-4(2x-y)+8

3(2x-y)(2x-y-2)-4(2x-y-2)

(2x-y-2)(6x-3y-40

\(\left(12x^2-12xy+3y^2\right)-10\left(2x-y\right)+8\)

\(=\left(12x^2-6xy-6xy+3y^2\right)-10\left(2x-y\right)+8\)

\(=\left[6x\left(2x-y\right)-3y\left(2x-y\right)\right]-10\left(2x-y\right)+8\)

\(=\left(2x-y\right)\left(6x-3y\right)-10\left(2x-y\right)+8\)

\(=3\left(2x-y\right)^2-10\left(2x-y\right)+8\)

Đặt \(2x-y=a\), khi đó biểu thức có dạng:

\(3a^2-10a+8=3a^2-6a-4a+8\)

\(=3a\left(a-2\right)-4\left(a-2\right)=\left(a-2\right)\left(3a-4\right)\)

\(=\left(2x-y-2\right)\left(6x-3y-4\right).\)

a) \(A=x^2-2xy+y^2+3x-3y-4\)

\(=\left(x-y\right)^2-1+3x-3y-3\)

\(=\left(x-y-1\right)\left(x-y+1\right)+3\left(x-y-1\right)\)

\(=\left(x-y-1\right)\left(x-y+1+3\right)\)

\(=\left(x-y-1\right)\left(x-y+4\right)\)

a,2x²-7x+6=(2x²-4x)-(3x-6)
=2x(x-3)-3(x-2)=(x-2)(2x-3)
b,x²+x-6=(x²+3x)-(2x+6)
=x(x-3)-2(x-3)=(x-3)(x-2)
c,x³+3x²+6x+4=x³+x²+2x²+2x+4x+4
=(x+1)(x²+2x+4)
d,x¹⁰+x⁵+1=(x¹⁰-x)+(x⁵-x²)+(x²+x+1)
=x((x³)³-1)+x²(x³-1)+(x²+x+1)
=x(x³-1)(x⁶+x³+1)+x²(x-1)(x²+x+1)+(x²+x+1)
=x(x-1)(x²+x+1)+x²(x-1)(x²+x+1)+(x²+x+1)
(x²+x+1)(x²-x+x³-x²+1)
e,(12x²-12xy+3y²)-10x(2x-y)=3(4x²-4xy+y²)-10x(2x-y)
=3(2x-y)²-10x(2x-y)=(2x-y)(6x-3y-10x)=(2x-y)(-4x-3y)

phân tích đa thức thành nhân tử

a,2x^2-7x+6
b,x^2+x-6
c,x^3+3x^2+6x+4
d,x^10+x^5+1
e,(12x^2-12xy+3y^2)-10x(2x-y)

\(2x^2-7x+6\)

\(=2x^2-3x-4x+6\)

\(=x\left(2x-3\right)-2\left(2x-3\right)\)

\(=\left(x-2\right)\left(2x-3\right)\)

b: \(x^2-6x+xy-6y\)

\(=x\left(x-6\right)+y\left(x-6\right)\)

\(=\left(x-6\right)\left(x+y\right)\)

c: \(2x^2+2xy-x-y\)

\(=2x\left(x+y\right)-\left(x+y\right)\)

\(=\left(x+y\right)\left(2x-1\right)\)

e: \(x^3-3x^2+3x-1=\left(x-1\right)^3\)