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

phân tích theo hằng đẳng thức rồi rút gọn là ra thôi bạn

Ta có: \(2sin2x+\sqrt{2}sin4x=0\)

\(\Rightarrow2sin2x+2\sqrt{2}sin2xcos2x=0\)

\(\Rightarrow sin2x+\sqrt{2}sin2xcos2x=0\)

\(\Rightarrow sin2x\left(1+\sqrt{2}cos2x\right)=0\)

\(\Rightarrow\left[\begin{array}{nghiempt}sin2x=0\left(1\right)\\cos2x=-\frac{1}{\sqrt{2}}\left(2\right)\end{array}\right.\)

\(\left(1\right)\Rightarrow2x=k\pi\Rightarrow x=\frac{k\pi}{2}\)

\(\left(2\right)\Rightarrow cos2x=cos\frac{3\pi}{4}\Rightarrow\left[\begin{array}{nghiempt}2x=\frac{3\pi}{4}+k2\pi\left(3\right)\\2x=-\frac{3\pi}{4}+k2\pi\left(4\right)\end{array}\right.\)

\(\left(3\right)\Rightarrow x=\frac{3\pi}{8}+k\pi\)

\(\left(4\right)\Rightarrow x=-\frac{3\pi}{8}+k\pi\)

                               Vậy \(x=\left\{\frac{k\pi}{2};\frac{3\pi}{8}+k\pi;-\frac{3\pi}{8}+k\pi\right\}\)

Đặt ẩn đi