\(\left(x+2\right)\left(x+3\right)-2\sqrt[]{x^2+5x+3}=6\)
\(\Leftrightarrow x^2+5x+6-2\sqrt[]{x^2+5x+3}=6\left(1\right)\)
Đặt \(t=\sqrt[]{x^2+5x+3}\ge0\)
\(pt\left(1\right)\Leftrightarrow t^2+3-2t=6\)
\(\Leftrightarrow t^2-2t-3=0\) \(\left(a-b+c=0\right)\)
\(\Leftrightarrow\left[{}\begin{matrix}t=-1\left(loại\right)\\t=3\end{matrix}\right.\)
\(\Leftrightarrow t=3\)
\(\Leftrightarrow\sqrt[]{x^2+5x+3}=3\left(3>0\right)\)
\(\Leftrightarrow x^2+5x+3=9\)
\(\Leftrightarrow x^2+5x-6=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=1\\x=-6\end{matrix}\right.\) \(\left(a+b+c=0\right)\)
ĐKXĐ: x^2+5x+3>=0
=>\(\left[{}\begin{matrix}x< =\dfrac{-5-\sqrt{13}}{2}\\x>=\dfrac{-5+\sqrt{13}}{2}\end{matrix}\right.\)
\(\left(x+2\right)\left(x+3\right)-2\sqrt{x^2+5x+3}=6\)
\(\Leftrightarrow\left(x^2+5x+6\right)-2\sqrt{x^2+5x+3}-6=0\)
=>\(\left(x^2+5x\right)-2\sqrt{x^2+5x+3}=0\)
=>\(x^2+5x+3-2\sqrt{x^2+5x+3}-3=0\)
=>\(\left(\sqrt{x^2+5x+3}-3\right)\left(\sqrt{x^2+5x+3}+1\right)=0\)
=>\(x^2+5x+3=9\)
=>x^2+5x-6=0
=>(x+6)(x-1)=0
=>x=-6 hoặc x=1
