c) Đặt \(t=x^2+x+1\) thì
\(t\left(t+1\right)-12=t^2+t-12=\left(t-3\right)\left(t+4\right)\)
\(=\left(x^2+x-2\right)\left(x^2+x+5\right)=\left(x+2\right)\left(x-1\right)\left(x^2+x+5\right)\)
d) \(\left[\left(x+2\right)\left(x+5\right)\right]\left[\left(x+3\right)\left(x+4\right)\right]-24\)
\(=\left(x^2+7x+10\right)\left(x^2+7x+12\right)-24\)
Đặt \(t=x^2+7x+11\) thì
\(\left(t-1\right)\left(t+1\right)-24=t^2-1-24=t^2-25\)
\(=\left(t-5\right)\left(t+5\right)\)
\(=\left(x^2+7x+11-5\right)\left(x^2+7x+11+5\right)\)
\(=\left(x^2+7x+6\right)\left(x^2+7x+16\right)\)
\(=\left(x+1\right)\left(x+6\right)\left(x^2+7x+16\right)\)
Rồi nha bạn 
phân tích đa thức thành nhân tử
a) \(\left(x^2+x\right)^2-2\left(x^2+x\right)-15\)
\(\Leftrightarrow\left(x^2+x\right)^2-5\left(x^2+x\right)+3\left(x^2+x\right)-15\)
\(\Leftrightarrow\left(x^2+x\right)\left(x^2+x-5\right)+3\left(x^2+x-5\right)\)
\(\Leftrightarrow\left(x^2+x+3\right)\left(x^2+x-5\right)\)
b) \(x^2+2xy+y^2-x-y-12=0\)
\(\Leftrightarrow\left(x+y\right)^2-\left(x+y\right)-12=0\)
\(\Leftrightarrow\left(x+y\right)^2-4\left(x+y\right)+3\left(x+y\right)-12=0\)
\(\Leftrightarrow\left(x+y-4\right)\left(x+y+3\right)=0\)
