đưa x vào căn

=> cs 2 th:

thêm dấu - trc x hoặc ko

sau đó đặt x-1=t

thay vào giải pt là ra 

hok tốt

ĐK: \(x-\frac{1}{x}\ge0;x\ne0\)

Đặt \(\sqrt{x-\frac{1}{x}}=t\Rightarrow x-\frac{1}{x}=t^2\)

Theo đề bài ta có hệ: \(\hept{\begin{cases}\left(x-1\right)^2+xt=2\\x-\frac{1}{x}=t^2\end{cases}}\Leftrightarrow\hept{\begin{cases}x^2-2x-1=-xt\\x^2-1=xt^2\end{cases}}\)

Lấy pt dưới trừ pt trên vế với vế: \(2x=xt^2+xt\)

\(\Leftrightarrow x\left(t^2+t-2\right)=0\Leftrightarrow\orbr{\begin{cases}t=1\\t=-2\left(L\right)\end{cases}}\left(\text{vì }x\ne0\right)\)

....

P/s: Em ko chắc nha!

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

Biểu thức  \(A\)  có nghĩa khi  \(\hept{\begin{cases}\sqrt{x}+1\ne0;\text{ }x\ge0\\\sqrt{x}-1\ne0\end{cases}}\)  \(\Leftrightarrow\)  \(\hept{\begin{cases}x\ge0\\x\ne1\end{cases}}\)

Ta có:

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

\(A=\frac{x+\sqrt{x}-2\sqrt{x}+2-2\sqrt{x}-2}{x-1}=\frac{x-3\sqrt{x}}{x-1}\)

Vậy,  \(A=\frac{x-3\sqrt{x}}{x-1}\)

đề đúng hk bn

\(\frac{x+\sqrt{x}}{3\sqrt{x}-1}=6+2\sqrt{5}\)

\(\Leftrightarrow\)\(\frac{\sqrt{x}\left(\sqrt{x}-1\right)}{3\left(\sqrt{x}-1\right)}=2\left(3+\sqrt{5}\right)\)

\(\Leftrightarrow\frac{\sqrt{x}}{3}=2\left(3+\sqrt{5}\right)\)

\(\Leftrightarrow\sqrt{x}=6\left(3+\sqrt{5}\right)\)

\(\Leftrightarrow x=36\left(3+\sqrt{5}\right)^2\)

\(\Leftrightarrow x=36\left(14+6\sqrt{5}\right)\)

ĐKXĐ là x\(\ge\frac{2}{3}\)

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

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

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

ĐKXĐ : x > 1

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

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

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

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

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

Để A = 9/2

=> \(\frac{x}{\sqrt{x}-1}=\frac{9}{2}\)( ĐK : x > 1 )

<=> 2x = 9( √x - 1 )

<=> 2x = 9√x - 9

<=> 2x + 9 = 9√x (1)

Bình phương hai vế

(1) <=> 4x² + 36x + 81 = 81x

     <=> 4x² + 36x + 81 - 81x = 0

     <=> 4x² - 45x + 81 = 0

     <=> 4x² - 36x - 9x + 81 = 0

     <=> 4x( x - 9 ) - 9( x - 9 ) = 0

     <=> ( x - 9 )( 4x - 9 ) = 0

     <=> \(\orbr{\begin{cases}x-9=0\\4x-9=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=9\\x=\frac{9}{4}\end{cases}}\)( tm )

a).  \(\frac{1}{\sqrt{5-\sqrt{7}}}+\frac{\sqrt{5}}{\sqrt{5+\sqrt{7}}})-1\)

\(\Leftrightarrow\frac{1}{\sqrt{25-\sqrt{49}}}-1\)

\(\Leftrightarrow\frac{1}{\sqrt{25-7}}-1\)

\(\Leftrightarrow\frac{1}{\sqrt{18}}-1\)

\(\Leftrightarrow\frac{1}{3\sqrt{2}}-1\) 

ĐẾN ĐÂY BN QUY ĐỒNG LÀ ĐC

Ta có: \(A=\left(\frac{\sqrt{x}}{3+\sqrt{x}}+\frac{x+9}{9-x}\right).\left(\frac{3\sqrt{x}+1}{x-3\sqrt{x}}-\frac{1}{\sqrt{x}}\right)\)    (   ĐK: \(x\ne0,\)\(x\ne9,\)\(x\ge3\))

     \(\Leftrightarrow A=\frac{\sqrt{x}.\left(3-\sqrt{x}\right)+x+9}{\left(3+\sqrt{x}\right).\left(3-\sqrt{x}\right)}.\frac{3\sqrt{x}+1-\sqrt{x}+3}{\sqrt{x}.\left(\sqrt{x}-3\right)}\)

     \(\Leftrightarrow A=\frac{3\sqrt{x}-x+x+9}{\left(3+\sqrt{x}\right).\left(3-\sqrt{x}\right)}.\frac{2\sqrt{x}+4}{\sqrt{x}.\left(\sqrt{x}-3\right)}\)

     \(\Leftrightarrow A=\frac{3\sqrt{x}-9}{\left(3+\sqrt{x}\right).\left(3-\sqrt{x}\right)}.\frac{2\sqrt{x}+4}{\sqrt{x}.\left(\sqrt{x}-3\right)}\)

     \(\Leftrightarrow A=\frac{3\left(\sqrt{x}-3\right)}{\left(3+\sqrt{x}\right).\left(3-\sqrt{x}\right)}.\frac{2\sqrt{x}+4}{\sqrt{x}.\left(\sqrt{x}-3\right)}\)

     \(\Leftrightarrow A=\frac{3.\left(2\sqrt{x}+4\right)}{\left(9-x\right).\sqrt{x}}\)

     \(\Leftrightarrow A=\frac{6\sqrt{x}+12}{9\sqrt{x}-x}\)