(x+1)(x+4) - \(3\sqrt{x^2+5x+2}=6\)
\(\Leftrightarrow\)\(x^2+5x+4\)-\(3\sqrt{x^2+5x+2}=6\)
\(\Leftrightarrow\)\(x^2+5x+2\) - \(3\sqrt{x^2+5x+2}=4\)
Đặt \(x^2+5x+2\)= t (ĐK: t\(\ge\)0)
=> \(t-3\sqrt{t}=4\)
\(\Leftrightarrow t-2.\dfrac{3}{2}t+\dfrac{9}{4}-\dfrac{25}{4}=0\)
\(\Leftrightarrow\left(\sqrt{t}-\dfrac{3}{2}\right)^2-\dfrac{25}{4}=0\)
\(\Leftrightarrow\left(\sqrt{t}-4\right)\left(\sqrt{t}+1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}\sqrt{t}=4\\\sqrt{t}=-1\left(ktmđk\right)\end{matrix}\right.\Rightarrow t=16\)
Ta có: \(x^2+5x+2=16\)
\(\Leftrightarrow\)\(x^2+5x-14=0\)
\(\Leftrightarrow\)\(\left(x-2\right)\left(x+7\right)=0\)
\(\Leftrightarrow\)\(\left[{}\begin{matrix}x=2\\x=-7\end{matrix}\right.\)
Vậy.
