Question

Phân tích thành nhân tử:

a) (x+x)²+4(x²+x)-12

b) y²+4y-12

Answer 1

a, (x+x)^2+4 (x^2+x)-12=(2x)²+ 4x(x+1)-12

=(2x)²+ 2x.2(x+1) + (x+1)² - (x+1)²-12

=(2x+x+1)²-x²-2x-1-12=(3x+1)²-(x+1)²-12

=(3x+1-x-1)(3x+1+x+1)-12=2x(4x+2)-12

=4x(2x+1)-12=4(2x²+x-12)

b, y²+4y-12=y²+4y+4-16=(y+2)²-4²

=(y+2+4)(y+2-4)=(y+6)(y-2)

Answer 2

a ) \(\left(x+x\right)^2+4\left(x^2+x\right)-12\)

\(=4x^2+4x^2+4x-12\)

\(=8x^2+4x-12\)

\(=4\left(2x^2+x-3\right)\)

\(=4\left(2x^2-2x+3x-3\right)\)

\(=3\left[2x\left(x-1\right)+3\left(x-1\right)\right]\)

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

b ) \(y^2+4y-12=\left(y^2+4y+4\right)-16\)

\(=\left(y+2\right)^2-4^2=\left(y+2-4\right)\left(y+2+4\right)=\left(y-2\right)\left(y+6\right)\)