a,\(ĐK:x>0,x\ne1,x\ne4\)

\(A=\left[\dfrac{\sqrt{x}-\sqrt{x}+1}{\sqrt{x}\left(\sqrt{x}-1\right)}\right]:\left[\dfrac{x-1-x+4}{\left(\sqrt{x}-1\right)\left(\sqrt{x}-2\right)}\right]\)

\(A=\dfrac{1}{\sqrt{x}\left(\sqrt{x}-1\right)}.\dfrac{\left(\sqrt{x}-1\right)\left(\sqrt{x}-2\right)}{3}=\dfrac{\sqrt{x}-2}{3\sqrt{x}}\)

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

\(=>A=\dfrac{\sqrt{2}-3}{3\sqrt{2}-3}\)

a) ĐKXĐ: \(\left\{{}\begin{matrix}\sqrt{x}\ge0\\\sqrt{x}-1>0\\\sqrt{x}-2>0\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x\ge0\\x>1\\x>4\end{matrix}\right.\) \(\Leftrightarrow x>4\)

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

\(=\dfrac{\sqrt{x}-\left(\sqrt{x}-1\right)}{\sqrt{x}\left(\sqrt{x}-1\right)}:\dfrac{\left(x-1\right)-\left(x-4\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-1\right)}\)

\(=\dfrac{1}{\sqrt{x}\left(\sqrt{x}-1\right)}.\dfrac{\left(\sqrt{x}-2\right)\left(\sqrt{x}-1\right)}{3}\) 

\(=\dfrac{\sqrt{x}-2}{3\sqrt{x}}\)

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

Thay \(x=1\) vào \(A\), ta được:

\(A=\dfrac{\sqrt{1}-2}{3\sqrt{1}}=\dfrac{1-2}{3}=-\dfrac{1}{3}\)

a: ĐKXĐ: \(\left\{{}\begin{matrix}x>0\\x\notin\left\{1;4\right\}\end{matrix}\right.\)

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

\(=\dfrac{\sqrt{x}-\sqrt{x}+1}{\sqrt{x}\left(\sqrt{x}-1\right)}:\dfrac{x-1-x+4}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-1\right)}\)

\(=\dfrac{1}{\sqrt{x}\left(\sqrt{x}-1\right)}\cdot\dfrac{\left(\sqrt{x}-2\right)\left(\sqrt{x}-1\right)}{3}\)

\(=\dfrac{\sqrt{x}-2}{3\sqrt{x}}\)

tìm cả đk giúp mik vs

ĐKXĐ: \(x>0;x\ne1\)

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

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

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

b.

\(x=4+2\sqrt{3}=\left(\sqrt{3}+1\right)^2\Rightarrow\sqrt{x}=\sqrt{3}+1\)

\(\Rightarrow A=\dfrac{4+2\sqrt{3}}{\sqrt{3}+1-1}=\dfrac{4+2\sqrt{3}}{\sqrt{3}}=\dfrac{6+4\sqrt{3}}{3}\)

c.

Để \(\sqrt{A}\) xác định \(\Rightarrow\sqrt{x}-1>0\Rightarrow x>1\)

Ta có:

\(\sqrt{A}=\sqrt{\dfrac{x}{\sqrt{x}-1}}=\sqrt{\dfrac{x}{\sqrt{x}-1}-4+4}=\sqrt{\dfrac{\left(\sqrt{x}-2\right)^2}{\sqrt{x}-1}+4}\ge\sqrt{4}=2\)

Dấu "=" xảy ra khi \(\sqrt{x}-2=0\Rightarrow x=4\)

a)ĐKXĐ:\(\begin{cases}x\ge0\\2\sqrt{x}-2\ne0\\1-x\ne0\\\end{cases}\)

`<=>` \(\begin{cases}x\ge0\\x\ne1\\\end{cases}\)

`B=1/(2sqrtx-2)-1/(2sqrtx+2)+sqrtx/(1-x)`

`=1/(2(sqrtx-1))-1/(2(sqrtx+1))-sqrtx/(x-1)`

`=(sqrtx+1-(sqrtx-1)-2sqrtx)/(2(sqrtx-1)(sqrtx+1))`

`=(2-2sqrtx)/(2(sqrtx-1)(sqrtx+1))`

`=(2(1-sqrtx))/(2(sqrtx-1)(sqrtx+1))`

`=-1/(sqrtx+1)`

`b)x=3`

`=>B=(-1)/(sqrt3+1)`

`=(-(sqrt3-1))/(3-1)`

`=(1-sqrt3)/2`

`c)|A|=1/2`

`<=>|(-1)/(sqrtx+1)|=1/2`

`<=>|1/(sqrtx+1)|=1/2`

`<=>1/(sqrtx+1)=1/2` do `1>0,sqrtx+1>=1>0`

`<=>sqrtx+1=2`

`<=>sqrtx=1`

`<=>x=1` loại vì `x ne 1`.

a) ĐKXĐ: \(\left\{{}\begin{matrix}x\ge0\\x\ne1\end{matrix}\right.\)

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

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

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

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

b) Thay x=3 vào B, ta được:

\(B=\dfrac{-1}{\sqrt{3}+1}=\dfrac{-\sqrt{3}+1}{2}\)

c) Ta có: \(\left|A\right|=\dfrac{1}{2}\)

nên \(\left[{}\begin{matrix}A=\dfrac{1}{2}\\A=-\dfrac{1}{2}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}\dfrac{-1}{\sqrt{x}+1}=\dfrac{1}{2}\\\dfrac{-1}{\sqrt{x}+1}=\dfrac{-1}{2}\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}\sqrt{x}+1=-2\\\sqrt{x}+1=2\end{matrix}\right.\Leftrightarrow x=1\)(loại)

Lời giải:

ĐKXĐ: $x\geq 0; x\neq 1$

a)

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

\(=\frac{x+\sqrt{x}+1}{x+1}:\frac{x+1-2\sqrt{x}}{(\sqrt{x}-1)(x+1)}=\frac{x+\sqrt{x}+1}{x+1}.\frac{(\sqrt{x}-1)(x+1)}{(\sqrt{x}-1)^2}=\frac{x+\sqrt{x}+1}{\sqrt{x}-1}\)

b) 

\(A=7\Leftrightarrow x+\sqrt{x}+1=7(\sqrt{x}-1)\)

\(\Leftrightarrow x-6\sqrt{x}+8=0\Leftrightarrow (\sqrt{x}-2)(\sqrt{x}-4)=0\)

\(\Leftrightarrow \left[\begin{matrix} x=4\\ x=16\end{matrix}\right.\) (đều thỏa mãn)

c) 

\(x=2(2+\sqrt{3})=4+2\sqrt{3}=3+1+2\sqrt{3.1}=(\sqrt{3}+1)^2\Rightarrow \sqrt{x}=\sqrt{3}+1\)

\(\Rightarrow A=\frac{4+2\sqrt{3}+\sqrt{3}+1+1}{\sqrt{3}}=\frac{6+3\sqrt{3}}{\sqrt{3}}=3+2\sqrt{3}\)

d)

\(A< 1\Leftrightarrow \frac{x+\sqrt{x}+1}{\sqrt{x}-1}-1<0\Leftrightarrow \frac{x-2\sqrt{x}+2}{\sqrt{x}-1}<0\)

\(\Leftrightarrow \frac{(\sqrt{x}-1)^2+1}{\sqrt{x}-1}<0\Leftrightarrow \sqrt{x}-1< 0\Leftrightarrow 0\leq x< 1\)

\(=\dfrac{x-2\sqrt{x}+1}{\sqrt{x}\left(\sqrt{x}-1\right)}\left(dk:x\ne0,\pm1\right)\)

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

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

Vậy \(A=\dfrac{\sqrt{x}-1}{\sqrt{x}}\)

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

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

\(=\dfrac{x\sqrt{x}-2x+\sqrt{x}-x\sqrt{x}-2x-\sqrt{x}}{2\sqrt{x}}=\dfrac{-4x}{2\sqrt{x}}=-2\sqrt{x}\)

b: Để A>-6 thì \(2\sqrt{x}< 6\)

=>0<x<9 

Kết hợp ĐKXĐ, ta được:

0<x<9 và x<>1

\(a,\)

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

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

\(=\dfrac{x-1}{2\sqrt{x}}.\dfrac{-4x}{x-1}\)

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

Vậy \(A=-2\sqrt{x}\)

\(b,\)Đề \(A\ge-6\) thì \(-2\sqrt{x}\ge-6\) \(\Leftrightarrow\sqrt{x}\le3\) \(\Leftrightarrow x\le3^2\Leftrightarrow x\le9\)

Vậy \(x\le9\) thì \(A\ge-6\)

a: ĐKXĐ: \(\left\{{}\begin{matrix}x\ge0\\x\ne9\end{matrix}\right.\)

b: Ta có: \(P=\left(\dfrac{2\sqrt{x}}{\sqrt{x}+3}+\dfrac{\sqrt{x}}{\sqrt{x}-3}-\dfrac{3x+3}{x-9}\right):\left(\dfrac{2\sqrt{x}-2}{\sqrt{x}-3}-1\right)\)

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

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

\(=\dfrac{-3}{\sqrt{x}+3}\)

c: Thay \(x=4-2\sqrt{3}\) vào P, ta được:

\(P=\dfrac{-3}{\sqrt{3}-1+3}=\dfrac{-3}{2+\sqrt{3}}=-6+3\sqrt{3}\)

a: Để P nguyên thì \(-3⋮\sqrt{x}+3\)

\(\Leftrightarrow\sqrt{x}+3=3\)

hay x=0