ĐK : \(\orbr{\begin{cases}2-x\ge0\\x^2-4\ge0\end{cases}}\Leftrightarrow\hept{\begin{cases}x\le-2\\x=2\end{cases}}\)

\(\sqrt{2-x}-\sqrt{x^2-4}=0\)

\(\Leftrightarrow\sqrt{2-x}=\sqrt{x^2-4}\)

\(\Leftrightarrow2-x=x^2-4\)

\(\Leftrightarrow x^2+x-6=0\)

\(\Leftrightarrow\orbr{\begin{cases}x=2\left(TM\right)\\x=-3\left(TM\right)\end{cases}}\)

Vậy \(S=\left\{2;-3\right\}\)

đề phải là \(-3x-3\sqrt{x}>0\)

\(-3\left(x+\sqrt{x}\right)\)kết hợp đk của x 

\(0< x< 1\)

\(< =>x+\sqrt{x}>0\)

\(< =>-3\left(x+\sqrt{x}\right)< 0\)

\(-3x-3\sqrt{x}< 0\)

\(6x\sqrt{2x^3+7}=6x^3+2x+22-4\sqrt{2x^3+7}\left(1\right)\) ĐK: \(\sqrt{2x^3+7}\ge0\)

Đặt \(\sqrt{2x^3+7}=a\ge0\)

\(\Rightarrow3a^2=6x^3+21\)

\(\Rightarrow\left(1\right)\Leftrightarrow6ax=3a^2+2x+1-4a\)

\(\Rightarrow\left(1\right)\Leftrightarrow3a^2+2x+1-4a-6ax=0\)

\(\Leftrightarrow\left(3a^2-4a+1\right)+2x\left(1-3a\right)=0\)

\(\Leftrightarrow\left(a-1\right)\left(3a-1\right)+2x\left(1-3a\right)=0\)

\(\Leftrightarrow\left(3a-1\right)\left(a-1-2x\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}a=\frac{1}{3}\\a=1+2x\end{cases}}\)

TH1: \(a=\frac{1}{3}\)

\(\Rightarrow\sqrt{2x^3+7}=\frac{1}{3}\)

\(\Leftrightarrow x^3=\frac{-31}{9}\)

\(\Leftrightarrow x=\sqrt[3]{\frac{-31}{9}}\left(tm\right)\)

TH2: a=1+2x

\(\Rightarrow\sqrt{2x^3+7}=1+2x\left(x\ge\frac{-1}{2}\right)\)

\(\Leftrightarrow2x^3+7=4x^2+4x+1\)

\(\Leftrightarrow x^3-2x^2-2x+3=0\)

\(\Leftrightarrow x^3-x^2-x^2+x-3x+3=0\)

\(\Leftrightarrow x^2\left(x-1\right)-x\left(x-1\right)-3\left(x-1\right)=0\)

\(\Leftrightarrow\left(x-1\right)\left(x^2-x-3\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}x=1\left(tm\right)\\x^2-x-3=0\left(2\right)\end{cases}}\)

Ta có: \(\Delta\left(2\right)=13\)

=> pt có 2 nghiệm pb \(\orbr{\begin{cases}x=\frac{1+\sqrt{13}}{2}\left(tm\right)\\x=\frac{1-\sqrt{13}}{2}\left(loai\right)\end{cases}}\)

Vậy ...

\(\sqrt{\frac{2x-3}{x-1}}=2\)

\(\Rightarrow\frac{2x-3}{x-1}=2^2\)

\(\Rightarrow\frac{2.x-3}{x-1}=4\)

\(\Rightarrow2.x-3=4.\left(x-1\right)\)

\(\Rightarrow2.x-3=4.x-4\)

\(\Rightarrow2.x=4.x-1\)

\(\Rightarrow4.x-2.x=1\)

\(\Rightarrow2.x=1\)

\(\Rightarrow x=\frac{1}{2}\)