ĐKXĐ: x>=3/2
\(\dfrac{3x-2}{\sqrt{x+1}}=\sqrt{x+1}+\sqrt{2x-3}\)
=>\(3x-2=x+1+\sqrt{\left(2x-3\right)\left(x+1\right)}\)
=>\(\sqrt{\left(2x-3\right)\left(x+1\right)}=2x-3\)
=>\(\sqrt{\left(2x-3\right)\left(x+1\right)}-\sqrt{\left(2x-3\right)\left(2x-3\right)}=0\)
=>\(\sqrt{\left(2x-3\right)}\cdot\left(\sqrt{x+1}-\sqrt{2x-3}\right)=0\)
=>2x-3=0 hoặc x+1=2x-3
=>-x=-4 hoặc x=3/2
=>x=3/2 hoặc x=4
\(\dfrac{3x-2}{\sqrt{x+1}}=\sqrt{x+1}+\sqrt{2x-3}\)
ĐKXĐ: \(x\ge\dfrac{3}{2}\)
\(\Leftrightarrow3x-2=\sqrt{x+1}\cdot\left(\sqrt{x+1}+\sqrt{2x-3}\right)\)
\(\Leftrightarrow3x-2=\sqrt{x+1}\cdot\sqrt{x+1}+\sqrt{x+1}\cdot\sqrt{2x-3}\)
\(\Leftrightarrow3x-2=\left(\sqrt{x+1}\right)^2+\sqrt{\left(x+1\right)\left(2x-3\right)}\)
\(\Leftrightarrow3x-2=x+1+\sqrt{\left(x+1\right)\left(2x-3\right)}\)
\(\Leftrightarrow3x-2-\left(x+1\right)=\sqrt{\left(x+1\right)\left(2x-3\right)}\)
\(\Leftrightarrow3x-2-x-1=\sqrt{\left(x+1\right)\left(2x-3\right)}\)
\(\Leftrightarrow2x-3=\sqrt{\left(x+1\right)\left(2x-3\right)}\)
\(\Leftrightarrow\sqrt{2x-3}\cdot\sqrt{2x-3}-\sqrt{\left(x+1\right)\left(2x-3\right)}=0\)
\(\Leftrightarrow\sqrt{2x-3}\cdot\sqrt{2x-3}-\sqrt{x+1}\cdot\sqrt{2x-3}=0\)
\(\Leftrightarrow\sqrt{2x-3}\cdot\left(\sqrt{2x-3}-\sqrt{x+1}\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}\sqrt{2x-3}=0\\\sqrt{2x-3}-\sqrt{x+1}=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}2x=3\\\sqrt{2x-3}=\sqrt{x+1}\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{3}{2}\left(tm\right)\\2x-3=x+1\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{3}{2}\left(tm\right)\\2x-x=1+3\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{3}{2}\left(tm\right)\\x=4\left(tm\right)\end{matrix}\right.\)
Vậy: \(S=\left\{\dfrac{3}{2};4\right\}\)
