mình giải thế này

a)\(P=\left(\frac{\left(\sqrt{x}-2\right)\left(\sqrt{x}+1\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)^2}-\frac{\left(\sqrt{x}+2\right)\left(\sqrt{x}-1\right)}{\left(\sqrt{x}+1\right)^2\left(\sqrt{x}-1\right)}\right)\frac{\left(1-x\right)^2}{2}\)

\(P=\frac{-2\sqrt{x}}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)^2}.\frac{\left(\sqrt{x}-1\right)^2\left(\sqrt{x+1}\right)^2}{2}\)

\(P=-\sqrt{x}.\left(\sqrt{x}-1\right)=-x+\sqrt{x}\)

b)\(0< x< 1\Rightarrow\sqrt{x}< 1\Rightarrow\sqrt{x}-1< 0\)

\(\Rightarrow-x\left(\sqrt{x}-1\right)>0\)vì \(x>0\)

xong rồi nhé :)

Hình như kết quả rút gọn là  \(\sqrt{x}-x\)

ĐKXĐ: \(x\ge4\)

a/ \(A=\frac{\sqrt{x}-\sqrt{x}+1}{\sqrt{x}\left(\sqrt{x}-1\right)}:\left[\frac{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)-\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}-2\right)}\right]\)

     \(=\frac{1}{\sqrt{x}\left(\sqrt{x}-1\right)}:\left(\frac{x-4-x+1}{\left(\sqrt{x}-1\right)\left(\sqrt{x}-2\right)}\right)\)

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

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

b/ A = 0 \(\Rightarrow\frac{\sqrt{x}-2}{-3\sqrt{x}}=0\Rightarrow\sqrt{x}-2=0\Rightarrow\sqrt{x}=2\Rightarrow x=4\)

Cho mình sửa lại:

Điều kiện: x > 4

nên câu b loại x = 4 nha

Answer:

a. \(P=\left(\frac{\sqrt{x}-2}{x-1}-\frac{\sqrt{x}+2}{x+2\sqrt{x}+1}\right)\left(\frac{1-x}{\sqrt{2}}\right)^2\)   ĐK: \(x\ge0;x\ne1\)

\(=\frac{\left(\sqrt{x}-2\right)\left(\sqrt{x}+1\right)-\left(\sqrt{x}+2\right)\left(\sqrt{x}-1\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)^2}.\frac{\left(1-x\right)^2}{2}\)

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

\(=\frac{\sqrt{x}\left(1-x\right)}{\sqrt{x}+1}\)

\(=\frac{\sqrt{x}\left(1-\sqrt{x}\right)\left(\sqrt{x}+1\right)}{\sqrt{x}+1}\)

\(=\sqrt{x}\left(1-\sqrt{x}\right)\)

b. Vì \(0< x< 1\Rightarrow\hept{\begin{cases}\sqrt{x}\ge0\\1-\sqrt{x}>0\end{cases}}\Rightarrow\sqrt{x}\left(1-\sqrt{x}\right)>0\)

Do vậy \(\sqrt{x}\left(1-\sqrt{x}\right)>0\)

c. \(P=\sqrt{x}\left(1-\sqrt{x}\right)\)

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

\(=-\left(x-2\sqrt{x}.\frac{1}{2}+\frac{1}{4}\right)+\frac{1}{4}\)

\(=-\left(\sqrt{x}-\frac{1}{2}\right)^2+\frac{1}{4}\le\frac{1}{4}\forall x\)

Dấu "=" xảy ra khi \(\sqrt{x}-\frac{1}{2}=0\Rightarrow x=\frac{1}{4}\)

K=\(\frac{\sqrt{x}+1}{\sqrt{x}+3}+\frac{\sqrt{x}-2}{\sqrt{x}-1}-\frac{2x-10}{x+2\sqrt{x}-3}ĐK:\hept{\begin{cases}x\ge0\\x\ne1\end{cases}}\)

=\(\frac{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)+\left(\sqrt{x}+2\right)\left(\sqrt{x}+3\right)-2x+10}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-1\right)}\)

=\(\frac{x-1-2x+3\sqrt{x}-2\sqrt{x}-1-6+10}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-1\right)}\)

=\(\frac{\sqrt{x}+3}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-1\right)}=\frac{1}{\sqrt{x}-1}\)

Để K>0 thì :\(\frac{1}{\sqrt{x}-1}>0\Leftrightarrow\sqrt{x}-1>0\Leftrightarrow x>1\)

Với x>1 thoả mãn yêu cầu.