Câu 1: 

\(B=\frac{\sqrt{x}}{\sqrt{x}-2}+\frac{4\sqrt{x}}{4-x}-\frac{2}{\sqrt{x}+2}\)

\(=\frac{\sqrt{x}\left(\sqrt{x}+2\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}-\frac{4\sqrt{x}}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}-\frac{2\left(\sqrt{x}-2\right)}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}\)

\(=\frac{x+2\sqrt{x}-4\sqrt{x}-2\sqrt{x}+4}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\)

\(=\frac{\left(\sqrt{x}-2\right)^2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\)

\(=\frac{\sqrt{x}-2}{\sqrt{x}+2}\)

\(P=A.B=\left(\frac{\sqrt{x}-2}{\sqrt{x}+2}\right)^2\)

\(P< P^2\Leftrightarrow P\left(1-P\right)< 0\Leftrightarrow P>1\)(vì \(P>0\)) 

\(\left(\frac{\sqrt{x}-2}{\sqrt{x}+2}\right)^2>1\Leftrightarrow\orbr{\begin{cases}\frac{\sqrt{x}-2}{\sqrt{x}+2}>1\\\frac{\sqrt{x}-2}{\sqrt{x}+2}< -1\end{cases}}\Leftrightarrow\orbr{\begin{cases}\sqrt{x}-2>\sqrt{x}+2\\\sqrt{x}-2< -\sqrt{x}-2\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}0\sqrt{x}>4\left(vn\right)\\\sqrt{x}< 0\left(vn\right)\end{cases}}\)

Vậy không có giá trị nào của \(x\)thỏa mãn. 

ĐK \(x\ge\frac{3}{2}\) hoặc \(x< 1\)

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

\(\Rightarrow2x-3=2x-2\Leftrightarrow0x=1\)ko có giá trị nào thoả mãn

Chúc học tốt!

Đk \(\orbr{\begin{cases}x\ge\frac{3}{2}\\x< 1\end{cases}}\)

\(\Leftrightarrow\frac{2x-3}{x-1}=4\)\(\Rightarrow2x-3=4x-4\Leftrightarrow2x=1\Leftrightarrow x=\frac{1}{2}\)TMĐK

Vậy x=1/2

Chúc học tốt! ( sửa lại)