theo mình thì:

/x-\(\sqrt{1-2x+x^2}\) / = /x-/x-1//=/x-x+1/(vì x>\(\sqrt{2}\) => x-1>0) = /1/=1

`(\sqrt(3x^2-12x+12)-x+2)/(x-2)`

`=(\sqrt(3(x^2-4x+4))-(x-2))/(x-2)`

`=(\sqrt(3(x-2)^2)) -(x-2))/(x-2)`

`=(\sqrt3. (x-2) - (x-2))/(x-2)`

`=( (\sqrt3-1) (x-2))/(x-2)`

`=\sqrt3-1`

`=>` C.

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

`->A`

\(=2x^2\cdot\dfrac{3}{x^2}=6\)

Rút gọn biểu thức 

c

B

\(a,C=\dfrac{2x^2-x-x-1+2-x^2}{x-1}\left(x\ne1\right)\\ C=\dfrac{x^2-2x+1}{x-1}=\dfrac{\left(x-1\right)^2}{x-1}=x-1\\ b,D=\dfrac{1+\sqrt{a}}{\sqrt{a}\left(\sqrt{a}-1\right)}\cdot\dfrac{\left(\sqrt{a}-1\right)^2}{\sqrt{a}+1}\left(a>0;a\ne1\right)\\ D=\dfrac{\sqrt{a}-1}{\sqrt{a}}\)

Có 

\(1,=0,9\left|x\right|\\ 2,Sửa:\dfrac{\sqrt{63y^3}}{\sqrt{7y}}=\sqrt{\dfrac{63y^3}{7y}}=\sqrt{9y^2}=3\left|y\right|=-3y\)

a, \(B=\dfrac{x}{\sqrt{x}-1}-\dfrac{2x-\sqrt{x}}{x-\sqrt{x}}\)ĐK : \(x>0;x\ne1\)

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

b,Ta có  \(x=3+2\sqrt{2}=\left(\sqrt{2}+1\right)^2\)

\(\Rightarrow\sqrt{x}=\sqrt{2}+1\)

Vậy \(B=\sqrt{2}+1-1=\sqrt{2}\)

a) Ta có: \(B=\dfrac{x}{\sqrt{x}-1}-\dfrac{2x-\sqrt{x}}{x-\sqrt{x}}\)

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

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

b) Thay \(x=3+2\sqrt{2}\) vào B, ta được:

\(B=\sqrt{2}+1-1=\sqrt{2}\)

Câu 1:

Sửa đề: \(B=\left(\dfrac{x}{x+3\sqrt{x}}+\dfrac{1}{\sqrt{x}+3}\right):\left(1-\dfrac{2}{\sqrt{x}}+\dfrac{6}{x+3\sqrt{x}}\right)\)

Ta có: \(B=\left(\dfrac{x}{x+3\sqrt{x}}+\dfrac{1}{\sqrt{x}+3}\right):\left(1-\dfrac{2}{\sqrt{x}}+\dfrac{6}{x+3\sqrt{x}}\right)\)

\(=\left(\dfrac{x}{\sqrt{x}\left(\sqrt{x}+3\right)}+\dfrac{1}{\sqrt{x}+3}\right):\left(\dfrac{x+3\sqrt{x}-2\left(\sqrt{x}+3\right)+6}{\sqrt{x}\left(\sqrt{x}+3\right)}\right)\)

\(=\dfrac{\sqrt{x}+1}{\sqrt{x}+3}:\dfrac{x+3\sqrt{x}-2\sqrt{x}-6+6}{\sqrt{x}\left(\sqrt{x}+3\right)}\)

\(=\dfrac{\sqrt{x}+1}{\sqrt{x}+3}\cdot\dfrac{\sqrt{x}\left(\sqrt{x}+3\right)}{x+\sqrt{x}}\)

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

Câu 3: 

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

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

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

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

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