a: Ta có: \(x-3\sqrt{x+1}=-3\)
\(\Leftrightarrow\sqrt{9x+9}=x+3\)
\(\Leftrightarrow x^2+6x+9-9x-9=0\)
\(\Leftrightarrow x\left(x-3\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=3\end{matrix}\right.\)
a, ĐK: \(x\ge-1\)
\(x-3\sqrt{x+1}=-3\)
\(\Leftrightarrow x+3\left(1-\sqrt{x+1}\right)=0\)
\(\Leftrightarrow x-\dfrac{3x}{1+\sqrt{x+1}}=0\)
\(\Leftrightarrow x\left(1-\dfrac{3}{1+\sqrt{x+1}}\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\1+\sqrt{x+1}=3\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\left(tm\right)\\x=3\left(tm\right)\end{matrix}\right.\)
b, ĐK: \(0\le x\le2\)
\(\sqrt{2x}-1=\sqrt{2-x}\)
\(\Leftrightarrow\sqrt{2x}=\sqrt{2-x}+1\)
\(\Leftrightarrow2x=2-x+1+2\sqrt{2-x}\)
\(\Leftrightarrow2\sqrt{2-x}=3x-3\)
\(\Leftrightarrow\left\{{}\begin{matrix}8-4x=9x^2-18x+9\\3x-3\ge0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}9x^2-14x+1=0\\x\ge1\end{matrix}\right.\)
\(\Leftrightarrow x=\dfrac{7+2\sqrt{10}}{9}\left(tm\right)\)
