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/ Điều kiện xác định tự tìm nhé.

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

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

b/ \(\frac{a-1}{\sqrt{a}}=a-2\Leftrightarrow a-1=a\sqrt{a}-2\sqrt{a}\)

Đặt \(t=\sqrt{a},t>0\) thì : \(t^2-1=t^3-2t\Leftrightarrow t^3-t^2-2t+1=0\)

Giải pt trên để tìm a.

\(đkxđ\Leftrightarrow x\ge0;x\ne1;x\ne4\)

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

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

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

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

\(A< \frac{1}{6}\Rightarrow\frac{\sqrt{a}-2}{3\sqrt{a}}>\frac{1}{6}\Rightarrow\frac{\sqrt{a}-2}{3\sqrt{a}}-\frac{1}{6}>0\)

\(\Rightarrow\frac{2\left(\sqrt{a}-2\right)}{6\sqrt{a}}-\frac{\sqrt{a}}{6\sqrt{a}}>0\Rightarrow\frac{\sqrt{a}-4}{6\sqrt{a}}>0\)

Vì \(6\sqrt{a}>0\Rightarrow\sqrt{a}-4>0\Rightarrow\sqrt{a}>4\Rightarrow a>16\)

Vậy \(P>\frac{1}{6}\Leftrightarrow a>16\)

Tự làm đi easy quá mà :)))) không biết quy đồng mà rút gọn hay sao

M ngon m làm đi nói nhiều 

a, Để A xác định =>  \(\hept{\begin{cases}\sqrt{a}+1\ne0\\\sqrt{a}-1\ne0\\\sqrt{a}>0\end{cases}}\)

=>\(\hept{\begin{cases}a\ne1\\a>0\end{cases}}\)

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

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

=  -2

b, Vì giá trị của A không phụ thuộc vào giá trị của a=>với mọi giá trị của biến a thì A=-2 luôn bé hơn 0

a,

ĐK :a>0    ;    a  khác 1 , khác 4

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

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

\(Q=\frac{1}{\sqrt{a}\left(\sqrt{a}-1\right)}:\frac{3}{\left(\sqrt{a}-2\right)\left(\sqrt{a}-1\right)}\)

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

\(Q=\frac{\sqrt{a}-2}{3\sqrt{a}}\)

b,

để Q đạt Giá Trị dương

\(\Rightarrow Q>0\Leftrightarrow\frac{\sqrt{a}-2}{3\sqrt{a}}>0\)

có \(a>0\Leftrightarrow\sqrt{a}>0\Leftrightarrow3\sqrt{a}>0\)

Suy Ra : để Q dương thì \(\sqrt{a}-2>0\)

\(\Leftrightarrow a>4\)   Thỏa mãn ĐK :  a > 0   ;a  khác 1 , khác 4

a) \(ĐKXĐ:-1< a< 1\)

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

\(=\left(\frac{3}{\sqrt{1+a}}+\frac{\sqrt{1-a}.\sqrt{1+a}}{\sqrt{1+a}}\right):\left[\frac{3}{\sqrt{\left(1-a\right)\left(1+a\right)}}+\frac{\sqrt{\left(1-a\right)\left(1+a\right)}}{\sqrt{\left(1-a\right)\left(1+a\right)}}\right]\)

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

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

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

b) \(a=\frac{\sqrt{3}}{2+\sqrt{3}}\)\(\Rightarrow1-a=1-\frac{\sqrt{3}}{2+\sqrt{3}}=\frac{2+\sqrt{3}-\sqrt{3}}{2+\sqrt{3}}=\frac{2}{2+\sqrt{3}}\)

\(=\frac{2\left(2-\sqrt{3}\right)}{\left(2+\sqrt{3}\right)\left(2-\sqrt{3}\right)}=\frac{4-2\sqrt{3}}{4-3}=4-2\sqrt{3}\)

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

Thay \(1-a=\left(\sqrt{3}-1\right)^2\)vào biểu thức ta được:

\(B=\sqrt{\left(\sqrt{3}-1\right)^2}=\left|\sqrt{3}-1\right|=\sqrt{3}-1\)