\(\sqrt{3x-3}-\sqrt{5-x}=\sqrt{2x-4}\)
ĐKXĐ: \(\left\{{}\begin{matrix}3x-3\ge0\\5-x\ge0\\2x-4\ge0\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}x\ge1\\x\le5\\x\ge2\end{matrix}\right.\)\(\Leftrightarrow2\le x\le5\)
Pt \(\Leftrightarrow\sqrt{3x-3}=\sqrt{2x-4}+\sqrt{5-x}\)
\(\Leftrightarrow3x-3=2x-4+2\sqrt{\left(2x-4\right)\left(5-x\right)}+5-x\)
\(\Leftrightarrow2\sqrt{\left(2x-4\right)\left(5-x\right)}=3x-2x+x-3+4-5\)
\(\Leftrightarrow2\sqrt{\left(2x-4\right)\left(5-x\right)}=2x-4\)
\(\Leftrightarrow\sqrt{\left(2x-4\right)\left(5-x\right)}=x-2\)
\(\Leftrightarrow\left(2x-4\right)\left(5-x\right)=\left(x-2\right)^2\)
\(\Leftrightarrow-2x^2+14x-20=x^2-4x+4\)
\(\Leftrightarrow-3x^2+18x-24=0\)
\(\Leftrightarrow x^2-6x+8=0\)
\(\Leftrightarrow x^2-2x-4x+8=0\)
\(\Leftrightarrow x\left(x-2\right)-4\left(x-2\right)=0\)
\(\Leftrightarrow\left(x-2\right)\left(x-4\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-2=0\\x-4=0\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x=2\\x=4\end{matrix}\right.\)(tm)
Vậy....................................................
