Điều kiện: 4
\(x\ge\frac{1}{2}\)
Ta có:
\(x\left(\sqrt{2x-1}-3\right)=\frac{2\left(2x^2-7x-15\right)}{x^2-6x+13}\)
\(\Leftrightarrow x.\frac{2\left(x-5\right)}{\sqrt{2x-1}+3}=\frac{2\left(x-5\right)\left(2x+3\right)}{x^2-6x+13}\)
\(\Leftrightarrow2\left(x-5\right)\left(\frac{x}{\sqrt{2x-1}+3}-\frac{2x+3}{x^2-6x+13}\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x-5=0\\\frac{x}{\sqrt{2x-1}+3}-\frac{2x+3}{x^2-6x+13}\left(1\right)\end{cases}}\)
\(\left(1\right)\Leftrightarrow\frac{\left(x-3\right)+3}{\sqrt{2x-1}+3}-\frac{\left(2x-1\right)+4}{\left(x-3\right)^2+4}=0\)
Đặt \(\hept{\begin{cases}\left(x-3\right)=a\\\sqrt{2x-1}=b\ge0\end{cases}}\)
\(\Rightarrow\frac{a+3}{b+3}-\frac{b^2+4}{a^2+4}=0\)
Tới đây thì đơn giản rồi nhé
