\(\left(x^2+3x+2\right)\left(x^2+5x+6\right)=\left(x^2+2x+x+2\right)\left(x^2+2x+3x+6\right)\)
\(=\left(x+1\right)\left(x+2\right)\left(x+2\right)\left(x+3\right)=\left[\left(x+1\right)\left(x+3\right)\right]\left(x+2\right)^2\)
\(=\left(x^2+4x+3\right)\left(x^2+4x+4\right).\text{Đặt: }x^2+4x+3\Rightarrow a\left(a+1\right)=72\)
\(\text{cái này bạn giải ra được:}a=8\text{ hoặc }a=-9\text{ thấy:}a+1=\left(x+2\right)^2\ge0\Rightarrow a\ge-1\Rightarrow a=8\)
\(\Leftrightarrow\left(x+2\right)^2=9\Leftrightarrow\orbr{\begin{cases}x+2=3\\x+2=-3\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=1\\x=-5\end{cases}}\)
