Đề câu a hình như sai rồi. Tui nghĩ vậy mới đúng nè:
\(a,\left(x^2+x\right)^2+4\left(x^2+x\right)-12\)
Đặt: \(x^2+x=t\) ta có:
\(t^2+4t-12=t^2-2t+6t-12\)
\(=t\left(t-2\right)+6\left(t-2\right)=\left(t-2\right)\left(6+t\right)\)
Vì vậy: \(\left(x^2+x\right)^2+4\left(x^2+x\right)-12=\left(x^2+x-2\right)\left(x^2+x+6\right)\)
Ta phân tích tiếp: \(x^2+x-2=x^2-x+2x-2\)
\(=x\left(x-1\right)+2\left(x-1\right)\)
\(=\left(x+2\right)\left(x-1\right)\)
Cuối cùng ta có: \(\left(x^2+x\right)^2+4\left(x^2+x\right)-12=\left(x-1\right)\left(x+2\right)\left(x^2+x+6\right)\)
\(b,\left(x^2+x+1\right)\left(x^2+x+2\right)-12\)
\(=x^4+x^3+2x^2+x^3+x^2+2x+x^2+x+2-12\)
\(=x^4+2x^3+4x^2+3x-10\)
\(=x^4-x^3+3x^3-3x^2+7x^2-7x+10x-10\)
\(=x^3\left(x-1\right)+3x^2\left(x-1\right)+7x\left(x-1\right)+10\left(x-1\right)\)
\(=\left(x-1\right)\left(x^3+3x^2+7x+10\right)\)
\(=\left(x-1\right)\left(x^3+2x^2+x^2+2x+5x+10\right)\)
\(=\left(x-1\right)\left[x^2\left(x+2\right)+x\left(x+2\right)+5\left(x+2\right)\right]\)
\(=\left(x-1\right)\left(x+2\right)\left(x^2+x+5\right)\)
