\(\frac{a\sqrt{a}-1}{a-\sqrt{a}}-\frac{a\sqrt{a}+1}{a+\sqrt{a}}\)

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

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

\(=\frac{2a^2-2a}{a^2-a}\)

\(=2\)( 1 )

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

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

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

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

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

Cộng ( 1 ) và ( 2 ) lại thì ta được biểu thức ban đầu:

\(2+\frac{2\left(a+1\right)}{\sqrt{a}}\)

Câu b,c em chịu:((

P/S:e ko bt đúng hay sai đâu ạ

Mk giải nốt phần còn lại nha

sai thì thông cảm

\(2+\frac{2\left(a+1\right)}{\sqrt{a}}=7\Leftrightarrow2a+2=5\sqrt{a}\)

\(\Leftrightarrow2a-5\sqrt{a}+2=0\)

\(\Leftrightarrow\left(2\sqrt{a}-1\right)\left(\sqrt{a}-2\right)=0\Rightarrow\orbr{\begin{cases}a=\frac{1}{4}\\a=4\end{cases}}\)

\(2+\frac{2\left(a+1\right)}{\sqrt{a}}>6\)\(\Rightarrow2a+2>4\sqrt{a}\Rightarrow2\left(a+1-2\sqrt{a}\right)>0\)

\(\Leftrightarrow\left(a+1-2\sqrt{a}\right)>0\Leftrightarrow\left(\sqrt{a}-1\right)^2>0\)

\(\Leftrightarrow a\ne1;a\ge0\)

a/ \(\sqrt{ab}+\sqrt{a}-\sqrt{b}\)

b/ \(\sqrt{ab}+\sqrt{a}-\sqrt{b}=2\)

\(\Leftrightarrow\left(\sqrt{a}-1\right)\left(\sqrt{b}+1\right)=1\)

Xong rồi nhá

a) đkxđ: \(a>0;a\ne1\)

Ta có:

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

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

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

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

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

\(P=\frac{4a+2\sqrt{a}+2}{\left(\sqrt{a}+1\right)\sqrt{a}}\)

b) \(P=7\)

\(\Leftrightarrow\frac{4a+2\sqrt{a}+2}{\left(\sqrt{a}+1\right)\sqrt{a}}=7\)

\(\Leftrightarrow4a+2\sqrt{a}+2=7a+7\sqrt{a}\)

\(\Leftrightarrow3a+5\sqrt{a}-2=0\)

\(\Leftrightarrow\left(3a-\sqrt{a}\right)+\left(6\sqrt{a}-2\right)=0\)

\(\Leftrightarrow\left(3\sqrt{a}-1\right)\sqrt{a}+2\left(3\sqrt{a}-1\right)=0\)

\(\Leftrightarrow\left(3\sqrt{a}-1\right)\left(\sqrt{a}+2\right)=0\)

Mà \(\sqrt{a}+2\ge2\left(\forall a\right)\)

\(\Rightarrow3\sqrt{a}-1=0\Leftrightarrow3\sqrt{a}=1\)

\(\Leftrightarrow\sqrt{a}=\frac{1}{3}\Rightarrow a=\frac{1}{9}\)

\(đkxđ\Leftrightarrow\hept{\begin{cases}a\ge0\\a\ne1\end{cases}}\)

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

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

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

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

\(=\frac{\left(a-1\right)^2.-4\sqrt{a}}{4a\left(a-1\right)}=\frac{a-1}{\sqrt{a}}\)

\(b,A< 0\Rightarrow\frac{a-1}{\sqrt{a}}< 0\)

Mà \(\sqrt{a}\ge0\Rightarrow a-1\le0\Rightarrow a\le1\)

\(A=2\Rightarrow\frac{a-1}{\sqrt{a}}=2\)

\(\Rightarrow a-1=2\sqrt{a}\Rightarrow a-2\sqrt{a}-1=0\)

\(\Rightarrow a-2\sqrt{a}+1-2=0\)

\(\Rightarrow\left(\sqrt{a}-1\right)^2-\sqrt{2}^2=0\)

\(\Rightarrow\left(\sqrt{a}-1-\sqrt{2}\right)\left(\sqrt{a}-1+\sqrt{2}\right)=0\)

\(\Rightarrow\orbr{\begin{cases}\sqrt{a}=1+\sqrt{2}\\\sqrt{a}=1-\sqrt{2}\end{cases}\Rightarrow\orbr{\begin{cases}a=\left(1+\sqrt{2}\right)^2=3+2\sqrt{2}\\a=\left(1-\sqrt{2}\right)^2=3-2\sqrt{2}\end{cases}}}\)

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

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

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

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

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

\(=\frac{a-1}{4a}.\frac{-4\sqrt{a}.}{1}\)

\(=\frac{1-a}{\sqrt{a}}\)

1/ 

a/ ĐKXĐ: \(x\ge0\) và \(x\ne\frac{1}{9}\)

 b/  \(P=\left[\frac{\left(\sqrt{x}-1\right)\left(3\sqrt{x}+1\right)-\left(3\sqrt{x}-1\right)+8\sqrt{x}}{\left(3\sqrt{x}+1\right)\left(3\sqrt{x}-1\right)}\right]:\left(\frac{3\sqrt{x}+1-3\sqrt{x}+2}{3\sqrt{x}+1}\right)\)

    \(=\frac{3x-2\sqrt{x}-1-3\sqrt{x}+1+8\sqrt{x}}{\left(3\sqrt{x}+1\right)\left(3\sqrt{x}-1\right)}.\frac{3\sqrt{x}+1}{3}\)

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

c/ \(P=\frac{6}{5}\Rightarrow\frac{x+\sqrt{x}}{3\sqrt{x}-1}=\frac{6}{5}\Rightarrow6\left(3\sqrt{x}-1\right)=5\left(x+\sqrt{x}\right)\)

                  \(\Rightarrow5x-13\sqrt{x}+6=0\Rightarrow\left(5\sqrt{x}-3\right)\left(\sqrt{x}-2\right)=0\)

                   \(\Rightarrow\orbr{\begin{cases}\sqrt{x}=\frac{3}{5}\\\sqrt{x}=2\end{cases}\Rightarrow\orbr{\begin{cases}x=\frac{9}{25}\\x=4\end{cases}}}\)

                                                      Vậy x = 9/25 , x = 4

1) a) ĐKXĐ :  \(0\le x\ne\frac{1}{9}\)

b) \(P=\left(\frac{\sqrt{x}-1}{3\sqrt{x}-1}-\frac{1}{3\sqrt{x}+1}+\frac{8\sqrt{x}}{9x-1}\right):\left(1-\frac{3\sqrt{x}-2}{3\sqrt{x}+1}\right)\)

\(=\left[\frac{\left(\sqrt{x}-1\right)\left(3\sqrt{x}+1\right)}{\left(3\sqrt{x}-1\right)\left(3\sqrt{x}+1\right)}-\frac{3\sqrt{x}-1}{\left(3\sqrt{x}-1\right)\left(3\sqrt{x}+1\right)}+\frac{8\sqrt{x}}{\left(3\sqrt{x}-1\right)\left(3\sqrt{x}+1\right)}\right]:\frac{3\sqrt{x}+1-3\sqrt{x}+2}{3\sqrt{x}+1}\)

\(=\frac{3x-2\sqrt{x}-1-3\sqrt{x}+1+8\sqrt{x}}{\left(3\sqrt{x}-1\right)\left(3\sqrt{x}+1\right)}.\frac{3\sqrt{x}+1}{3}=\frac{3x+3\sqrt{x}}{3\left(3\sqrt{x}-1\right)}=\frac{x+\sqrt{x}}{3\sqrt{x}-1}\)

c) \(P=\frac{6}{5}\Leftrightarrow18\sqrt{x}-6=5x+5\sqrt{x}\Leftrightarrow5x-13\sqrt{x}+6=0\)

\(\Leftrightarrow\orbr{\begin{cases}x=\frac{9}{25}\\x=4\end{cases}}\)

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

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

b) \(19-8\sqrt{3}=\left(\sqrt{3}-4\right)^2\Rightarrow P=\sqrt{\left(\sqrt{3}-4\right)^2}-1=4-\sqrt{3}-1=3-\sqrt{3}\)

c) P < 1 <=> \(\sqrt{a}-1< 1\Leftrightarrow a< 4\)

Kết hợp với điều kiện : \(P< 1\Leftrightarrow\hept{\begin{cases}0< a< 4\\a\ne1\end{cases}}\)