\(Câu1\)
\(\left(x^2+x^2\right)+4x^2+4x-12=x^2+x^2+4x^2+4x-12\)
\(=6x^2+4x-12=2\cdot3x^2+2\cdot2x-2\cdot6\)
\(=2\left(3x^2+2x-6\right).\)
\(Câu2\)
\(\left(x-4\right)^2+\left(x-4\right)=\left(x-4\right)^2+x-4=x^2-8x+16+x-4\)
\(=x^2-8x+x+16-4=x^2-7x+12\)
\(=\left(x^2-3x\right)+\left(-4x+12\right)=x\left(x-3\right)-4\left(x-3\right)\)
\(=\left(x-3\right)\left(x-4\right).\)
Đúng 0
Bình luận (0)