ĐKXĐ: \(-\frac{1}{3}\le x< 2\)
\(2\sqrt{3x+1}-\frac{3}{\sqrt{2-x}}=3-2x\)
\(\Rightarrow2\sqrt{\left(3x+1\right)\left(2-x\right)}-3=\left(3-2x\right)\sqrt{2-x}\)
\(\Rightarrow2\sqrt{-3x^2+5x+2}-3=\left(3-2x\right)\sqrt{2-x}\)
\(\Rightarrow\left(2\sqrt{-3x^2+5x+2}-4\right)-\left(3-2x\right)\left(\sqrt{2-x}-1\right)+2x-2=0\)
\(\Rightarrow\frac{4\left(-3x^2+5x+2\right)-16}{2\sqrt{-3x^2+5x+2}+4}-\frac{\left(3-2x\right)\left(2-x-1\right)}{\sqrt{2-x}+1}+2\left(x-1\right)=0\)
\(\Rightarrow\frac{\left(x-1\right)\left(3x-2\right)}{2\sqrt{-3x^2+5x+2}+4}-\frac{\left(3-2x\right)\left(1-x\right)}{\sqrt{2-x}+1}+2\left(x-1\right)=0\)
\(\Rightarrow\left(x-1\right)\left(\frac{3x-2}{2\sqrt{-3x^2+5x+2}+4}+\frac{3-2x}{\sqrt{2-x}+1}+2\right)=0\)
\(\Rightarrow\left[\begin{array}{nghiempt}x=1\\\frac{3x-2}{2\sqrt{-3x^2+5x+2}+4}+\frac{3-2x}{\sqrt{2-x}+1}+2=0\end{array}\right.\)
Tới đây bạn giải tiếp nha
