Question

Giải PT sau: \(\sqrt{5x^2+10x+1}=7-x^2-2x\)

Answer 1

Đk: \(5x^2+10x+1\ge0\)

Đặt \(t=\sqrt{5x^2+10x+1}\ge0\)

\(pt\Leftrightarrow\sqrt{5x^2+10x+1}=\frac{-\left(5x^2+10x+1\right)}{5}+\frac{36}{5}\)

\(\Leftrightarrow5t=-t^2+36\Leftrightarrow t^2+5t-36=0\)

\(\Leftrightarrow\left(t-4\right)\left(t+9\right)=0\Leftrightarrow t=4\) ( do \(t\ge0\) )

\(\Leftrightarrow5x^2+10x+1=16\Leftrightarrow5x^2+10x-15=0\)

\(\Leftrightarrow5\left(x-1\right)\left(x+3\right)=0\Leftrightarrow\left[{}\begin{matrix}x=1\\x=-3\end{matrix}\right.\)( TM )