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

Mong mọi người giúp đỡ mình , mình đang cần gấp , cảm ơn mọi người 

Ta có HĐT : \(\hept{\begin{cases}a\sqrt{a}+b\sqrt{b}=\left(\sqrt{a}+\sqrt{b}\right)\left(a-\sqrt{ab}+b\right)\\a\sqrt{a}-b\sqrt{b}=\left(\sqrt{a}-\sqrt{b}\right)\left(a+\sqrt{ab}+b\right)\end{cases}\left(a,b\ge0\right)}\)

\(P=\left(\frac{2a+1}{a\sqrt{a}-1}-\frac{\sqrt{a}}{a+\sqrt{a}+1}\right)\times\left(\frac{1+\sqrt{a^3}}{1+\sqrt{a}}-\sqrt{a}\right)\)

ĐKXĐ : \(\hept{\begin{cases}a\ge0\\a\ne1\end{cases}}\)

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

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

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

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

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

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

b) \(P\times\sqrt{1-a}\)

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

ĐKXĐ: \(0\le x< 1\)

Với \(0\le x< 1\)

Ta có :\(\hept{\begin{cases}\sqrt{a}\le\sqrt{1}=1\Rightarrow\sqrt{a}-1\le0\\\sqrt{1-a}\ge0\end{cases}}\)

\(\Rightarrow\left(\sqrt{a}-1\right)\left(\sqrt{1-a}\right)\le0\)

Phần b là sao vậy tôi không hiểu lắm

Bài 1:

a)  \(\frac{2}{\sqrt{3}-1}-\frac{2}{\sqrt{3}+1}\)

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

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

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

b)   \(\frac{2}{5-\sqrt{3}}+\frac{3}{\sqrt{6}+\sqrt{3}}\)

\(=\frac{2\left(5+\sqrt{3}\right)}{\left(5-\sqrt{3}\right)\left(5+\sqrt{3}\right)}+\frac{3\left(\sqrt{6}-\sqrt{3}\right)}{\left(\sqrt{6}+\sqrt{3}\right)\left(\sqrt{6}-\sqrt{3}\right)}\)

\(=\frac{2\left(5+\sqrt{3}\right)}{2}+\frac{3\left(\sqrt{6}-\sqrt{3}\right)}{3}\)

\(=5+\sqrt{3}+\sqrt{6}-\sqrt{3}=5+\sqrt{6}\)

c)  ĐK:  \(a\ge0;a\ne1\)

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

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

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

\(=1-a+a=1\)

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 : \(P< 1\Leftrightarrow\hept{\begin{cases}0< a< 4\\a\ne1\end{cases}}\)

a,   A\(=\left(\frac{\left(\sqrt{x}+1\right)^2-\left(\sqrt{x}-1\right)^2+4\sqrt{x}\left(x-1\right)}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}\right):\frac{x-1}{\sqrt{x}}\)  ĐK  x>0   ;\(x\ne1;x\ne-1\)

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

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

b,  Để  A =2  \(\Rightarrow\frac{4x^2}{\left(x-1\right)^2}=2\Rightarrow4x^2=2\left(x-1\right)^2\)

                     <=>  \(4x^2=2x^2-4x+2\)

                      <=> \(2x^2+4x-2=0\)

                       <=> \(x^2+2x-1=0\)

                       \(\Delta=1^2-1.\left(-1\right)\) =  2

                => \(\orbr{\begin{cases}x_1=-1-\sqrt{2}\left(loại\right)\\x_2=-1+\sqrt{2}\left(nhận\right)\end{cases}}\)

Vậy x=\(-1+\sqrt{2}\)thì  A =2  

c, Thay   x =\(\left(4+\sqrt{15}\right)\left(\sqrt{10}-\sqrt{6}\right)\sqrt{4-\sqrt{15}}\)=2

  =>A  =   \(\frac{4.2^2}{\left(2-1\right)^2}=16\)

Vậy  A=16  thì  x=\(\left(4+\sqrt{15}\right)\left(\sqrt{10}-\sqrt{6}\right)\sqrt{4-\sqrt{15}}\)