\(a,ĐK:x\ge-2\\ PT\Leftrightarrow3\sqrt{x+2}=5x-4\\ \Leftrightarrow9\left(x+2\right)=25x^2-40x+16\\ \Leftrightarrow25x^2-49x-2=0\\ \Leftrightarrow\left(x-2\right)\left(25x+1\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=2\left(tm\right)\\x=-\dfrac{1}{25}\left(tm\right)\end{matrix}\right.\\ b,ĐK:-1\le x\le5\\ PT\Leftrightarrow\left[{}\begin{matrix}\sqrt{x+3}=2\\\sqrt{5-x}=\sqrt{x+1}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x+3=4\\5-x=x+1\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=1\left(tm\right)\\x=2\left(tm\right)\end{matrix}\right.\)
\(c,ĐK:x\ge5\\ PT\Leftrightarrow\left[{}\begin{matrix}\sqrt{x-5}=-2\left(ktm\right)\\\left|x+1\right|=8-x\left(1\right)\end{matrix}\right.\\ \left(1\right)\Leftrightarrow\left[{}\begin{matrix}x+1=8-x,\forall x\ge-1\\x+1=x-8,\forall x< -1\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=\dfrac{7}{2}\left(tm\right)\\0x=-9\left(ktm\right)\end{matrix}\right.\Leftrightarrow x=\dfrac{7}{2}\)
Vậy pt có nghiệm \(x=\dfrac{7}{2}\)
