\(\left(x+1\right)\left(x+4\right)=5\sqrt{x^2+5x+28}\)
\(Pt\Leftrightarrow x^2+5x+4-5\sqrt{x^2+5x+28}=0\)
Đặt \(t=\sqrt{x^2+5x+28}\) \(\left(t>0\right)\)
Ta có: \(t^2=x^2+5x+28\)
\(\Leftrightarrow x^2+5x+4=t^2-24\)
Thay vào pt ta được:
\(t^2-24-5t=0\)
\(\Leftrightarrow t^2-8t+3t-24=0\)
\(\Leftrightarrow\left(t-8\right)\left(t+3\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}t=8\left(tm\right)\\t=-3\left(loai\right)\end{matrix}\right.\)
*) \(t=8\Leftrightarrow x^2+5x-36=0\)
\(\Leftrightarrow x^2+9x-4x-36=0\)
\(\Leftrightarrow x\left(x+9\right)-4\left(x+9\right)=0\)
\(\Leftrightarrow\left(x+9\right)\left(x-4\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-9\\x=4\end{matrix}\right.\)(tm)
Vậy.................