ĐKXĐ: \(\left[{}\begin{matrix}x\ge-1\\x\le-3\end{matrix}\right.\)
Ta có: \(\sqrt{x+3}+2x\sqrt{x+1}=2x+\sqrt{x^2+4x+3}\)
\(\Leftrightarrow\sqrt{x+3}+2x\sqrt{x+1}-2x-\sqrt{\left(x+1\right)\left(x+3\right)}=0\)
\(\Leftrightarrow\sqrt{x+3}\left(1-\sqrt{x+1}\right)-2x\left(1-\sqrt{x+1}\right)=0\)
\(\Leftrightarrow\left(1-\sqrt{x+1}\right)\left(\sqrt{x+3}-2x\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x+1=1\\x+3=4x^2\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\4x^2-x-3=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=1\\x=\dfrac{-3}{4}\end{matrix}\right.\)
Đúng 2
Bình luận (1)
