a, \(P=\left(\frac{x\sqrt{x}}{\sqrt{x}+1}+\frac{x^2}{x\sqrt{x}+1}\right)\left(2-\frac{1}{\sqrt{x}}\right)\)ĐK : \(x\ge0;\sqrt{x}+1>0\)

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

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

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

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

b, \(P=0\Rightarrow\frac{x\left(x+1\right)\left(2\sqrt{x}-1\right)}{\left(\sqrt{x}+1\right)\left(x-\sqrt{x}+1\right)}=0\Leftrightarrow x\left(x+1\right)\left(2\sqrt{x}-1\right)=0\)

\(\Leftrightarrow x=0;x=-1;x=\frac{1}{4}\)Kết hợp với đk vậy \(x=0;x=\frac{1}{4}\)

\(P=\frac{x^2-\sqrt{x}}{x+\sqrt{x}+1}-\frac{2x+\sqrt{x}}{\sqrt{x}}+\frac{2'x-1'}{\sqrt{x}-1}\) 

Rút gọn ta được:

\(P=\frac{x^1-\sqrt{x}}{x+\sqrt{x}+1}-\frac{1x+\sqrt{x}}{\sqrt{x}}+\frac{1'x-1'}{\sqrt{x}-1}\)

Phần \(\frac{2'x-1'}{\sqrt{x-1}}\) rút gọi được phần 2 thôi

Đề không yêu cầu Giải Phương trình nhé :v

P/s: Có chắc không nhỉ ?

mình không hiểu bạn làm cho lắm?

a: \(A=\dfrac{\left(\sqrt{a}-\sqrt{b}\right)^2}{\sqrt{a}-\sqrt{b}}-\dfrac{\sqrt{ab}\left(\sqrt{a}+\sqrt{b}\right)}{\sqrt{ab}}\)

\(=\sqrt{a}-\sqrt{b}-\sqrt{a}-\sqrt{b}=-2\sqrt{b}\)

b: \(B=\dfrac{2\sqrt{x}-x-x-\sqrt{x}-1}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}\cdot\dfrac{x+\sqrt{x}+1}{x-1}\)

\(=\dfrac{-2x+\sqrt{x}-1}{\sqrt{x}-1}\cdot\dfrac{1}{x-1}\)

c: \(C=\dfrac{x-9-x+3\sqrt{x}}{x-9}:\left(\dfrac{3-\sqrt{x}}{\sqrt{x}-2}+\dfrac{\sqrt{x}-2}{\sqrt{x}+3}+\dfrac{x-9}{x+\sqrt{x}-6}\right)\)

\(=\dfrac{3\left(\sqrt{x}-3\right)}{x-9}:\dfrac{9-x+x-4\sqrt{x}+4+x-9}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-2\right)}\)

\(=\dfrac{3}{\sqrt{x}+3}\cdot\dfrac{\left(\sqrt{x}+3\right)\left(\sqrt{x}-2\right)}{x-4\sqrt{x}+4}\)

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

Bài 1 : Với : \(x>0;x\ne1\)

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

Thay vào ta được : \(P=x=25\)

Bài 2 : 

a, Với \(x\ge0;x\ne1\)

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

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

Thay x = 9 vào A ta được : \(\frac{3}{3+1}=\frac{3}{4}\)

Bài 3 : \(x\ge0;x\ne1\)

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

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

b, Ta có : \(P=\frac{\sqrt{x}+2}{\sqrt{x}-1}=\frac{5}{4}\Rightarrow4\sqrt{x}+8=5\sqrt{x}-5\)

\(\Leftrightarrow\sqrt{x}=13\Leftrightarrow x=169\)(tmđk )

\(a,ĐKXĐ:x\ge0;x\ne1\)

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

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

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

\(P=\left(x+1\right)^2\left(x-1\right)^2\)

\(P=\left[\left(x+1\right)\left(x-1\right)\right]^2\)

\(P=\left(x^2+x-x-1\right)^2\)

\(P=\left(x^2-1\right)^2\)

b, \(7-4\sqrt{3}=2^2-4\sqrt{3}+\sqrt{3}\)

\(\left(2-\sqrt{3}\right)^2\)

\(P=\left(x^2-1\right)^2< \left(2-\sqrt{3}\right)^2\)

\(x^2-1< 2-\sqrt{3}\)

\(x^2< 3-\sqrt{3}\)

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

a) ĐKXĐ: \(\hept{\begin{cases}x\ge0\\1-\sqrt{x}\ne0\\1+\sqrt{x}\ne0\end{cases}}\) <=> \(\hept{\begin{cases}x\ge0\\x\ne1\end{cases}}\)

Ta có: \(P=\left(\frac{1-x\sqrt{x}}{1-\sqrt{x}}+\sqrt{x}\right)\left(\frac{1+x\sqrt{x}}{1+\sqrt{x}}-\sqrt{x}\right)\)

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

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

b) Với x > = 0 và x khác 1

Ta có: \(P< 7-4\sqrt{3}\)

<=> \(\left(x-1\right)^2< \left(2-\sqrt{3}\right)^2\)

<=> \(\left(x-1-2+\sqrt{3}\right)\left(x-1+2-\sqrt{3}\right)< 0\)

<=> \(\left(x-3+\sqrt{3}\right)\left(x+1-\sqrt{3}\right)< 0\)

<=> \(\hept{\begin{cases}x-3+\sqrt{3}< 0\\x+1-\sqrt{3}>0\end{cases}}\) hoặc \(\hept{\begin{cases}x-3+\sqrt{3}>0\\x+1-\sqrt{3}< 0\end{cases}}\)

<=> \(\hept{\begin{cases}x< 3-\sqrt{3}\\x>\sqrt{3}-1\end{cases}}\) hoặc \(\hept{\begin{cases}x>3-\sqrt{3}\\x< \sqrt{3}-1\end{cases}}\)

<=> \(\sqrt{3}-1< x< 3-\sqrt{3}\)