a) ĐK: x ≥ 0; x ≠ 9; x≠4

P= \(\left(\dfrac{\sqrt{x}+2}{\sqrt{x}-3}+\dfrac{3}{x-5\sqrt{x}+6}\right):\left(\dfrac{x+2}{\sqrt{x}-3}-\dfrac{x^2-\sqrt{x}-6}{\left(x-2\right)\left(\sqrt{x}-3\right)}\right)\)

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

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

=\(\dfrac{x-4+3}{\left(\sqrt{x}-3\right)\left(\sqrt{x}-2\right)}:\dfrac{x^2-4-x^2+\sqrt{x}+6}{\left(x-2\right)\left(\sqrt{x}-3\right)}\)

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

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

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

=\(\dfrac{x^2-3x+2}{x-4}\)

b)  P ≤ -2

⇒ \(\dfrac{x^2-3x+2}{x-4}\) ≤ -2

⇔ \(\dfrac{x^2-3x+2}{x-4}\) + 2 ≤ 0

⇔ \(\dfrac{x^2-3x+2+2\left(x-4\right)}{x-4}\) ≤ 0

⇔ \(\dfrac{x^2-3x+2+2x-8}{x-4}\) ≤ 0

⇔\(\dfrac{x^2-x-6}{x-4}\) ≤ 0

⇔ \(\left[{}\begin{matrix}\left\{{}\begin{matrix}x^2-x-6\ge0\\x-4< 0\end{matrix}\right.\\\left\{{}\begin{matrix}x^2-x-6\le0\\x-4>0\end{matrix}\right.\end{matrix}\right.\)

⇔\(\left[{}\begin{matrix}x\le2\\3\le x< 4\end{matrix}\right.\)

Vậy.......

`P=(sqrtx/(sqrtx-1)+sqrtx/(x-1)):(2/x-(2-x)/(xsqrtx+x))`

`đk:x>0,x ne 1`

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

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

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

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

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

`=x/(sqrtx-1)`

`b)P>2`

`<=>x/(sqrtx-1)-2>0`

`<=>(x-2sqrtx+2)/(sqrtx-1)>0`

`<=>((sqrtx-1)^2+1)/(sqrtx-1)>0`

`<=>sqrtx-1>0`

`<=>x>1`

a) đk: x>0;x khác 1;0

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

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

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

= \(\dfrac{x}{\sqrt{x}-1}\)

b) Để P > 2

<=> \(\dfrac{x}{\sqrt{x}-1}-2>0\)

<=> \(\dfrac{x-2\sqrt{x}+2}{\sqrt{x}-1}>0\)

<=> \(\dfrac{\left(\sqrt{x}-1\right)^2+1}{\sqrt{x}-1}>0\)

<=> \(\sqrt{x}-1>0\)

<=> x > 1

a) ĐKXĐ: \(x\ge0,x\ne1\)

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

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

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

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

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

\(=\dfrac{4\sqrt{x}}{\sqrt{x}+1}\)

b) \(P=\sqrt{x}-1\Rightarrow\dfrac{4\sqrt{x}}{\sqrt{x}+1}=\sqrt{x}-1\Rightarrow4\sqrt{x}=\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)\)

\(\Rightarrow4\sqrt{x}=x-1\Rightarrow x-4\sqrt{x}-1=0\)

\(\Delta=\left(-4\right)^2-4.\left(-1\right)=20\Rightarrow\left[{}\begin{matrix}\sqrt{x}=\dfrac{-b-\sqrt{\Delta}}{2a}=\dfrac{4-2\sqrt{5}}{2}=2-\sqrt{5}\\\sqrt{x}=\dfrac{-b+\sqrt{\Delta}}{2a}=\dfrac{4+2\sqrt{5}}{2}=2+\sqrt{5}\end{matrix}\right.\)

mà \(\sqrt{x}\ge0\Rightarrow\sqrt{x}=2+\sqrt{5}\Rightarrow x=9+4\sqrt{5}\)

c) \(P=\dfrac{4\sqrt{x}}{\sqrt{x}+1}=\dfrac{4\left(\sqrt{x}+1\right)-4}{\sqrt{x}+1}=4-\dfrac{4}{\sqrt{x}+1}\)

Để \(P\in Z\Rightarrow4⋮\sqrt{x}+1\Rightarrow\sqrt{x}+1\in\left\{1;2;4\right\}\left(\sqrt{x}+1\ge1\right)\)

\(\Rightarrow x\in\left\{0;1;9\right\}\) mà \(x\ne1\Rightarrow x\in\left\{0;9\right\}\)

giúp mình với ạ:<

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

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

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

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

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

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

b) Thay \(x=5+2\sqrt{3}\) vào biểu thức \(P=\dfrac{1}{x-1}\), ta được:

\(P=\dfrac{1}{5+2\sqrt{3}-1}=\dfrac{1}{4+2\sqrt{3}}\)

\(\Leftrightarrow P=\left(\dfrac{1}{\sqrt{3}+1}\right)^2\)

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

Vậy: Khi \(x=5+2\sqrt{3}\) thì \(\sqrt{P}=\dfrac{\sqrt{3}-1}{2}\)

a) \(P=\left(\dfrac{\sqrt{x}-1}{\sqrt{x}+1}-\dfrac{\sqrt{x}+1}{\sqrt{x}-1}\right).\left(\dfrac{1}{2\sqrt{x}}-\dfrac{\sqrt{x}}{2}\right)^2\left(đk:x>0\right)\)

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

b) \(P-\left(-2\sqrt{x}\right)=\dfrac{1-x}{\sqrt{x}}+2\sqrt{x}=\dfrac{1-x+2x}{\sqrt{x}}=\dfrac{1+x}{\sqrt{x}}>0\)

\(\Rightarrow P>-2\sqrt{x}\)

a, ĐK: \(x\ge0;x\ne1\)

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

\(=\dfrac{\left(\sqrt{x}-1\right)^2-\left(\sqrt{x}+1\right)^2}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}.\dfrac{\left(2-2x\right)^2}{16x}\)

\(=\dfrac{-4\sqrt{x}}{x-1}.\dfrac{4\left(x-1\right)^2}{16x}\)

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

a: Ta có: \(P=\left(\dfrac{\sqrt{x}-1}{\sqrt{x}+1}-\dfrac{\sqrt{x}+1}{\sqrt{x}-1}\right)\cdot\left(\dfrac{1}{2\sqrt{x}}-\dfrac{\sqrt{x}}{2}\right)^2\)

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

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

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

a) Ta có: \(A=\left(\dfrac{\sqrt{x}-4}{\sqrt{x}\left(\sqrt{x}-2\right)}+\dfrac{3}{\sqrt{x}-2}\right):\left(\dfrac{\sqrt{x}+2}{\sqrt{x}}-\dfrac{\sqrt{x}}{\sqrt{x}-2}\right)\)

\(=\dfrac{\sqrt{x}-4+3\sqrt{x}}{\sqrt{x}\left(\sqrt{x}-2\right)}:\dfrac{x-4-x}{\sqrt{x}\left(\sqrt{x}-2\right)}\)

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

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

b) Thay \(x=6-2\sqrt{5}\) vào A, ta được:

\(A=-\sqrt{6-2\sqrt{5}}+1=-\left(\sqrt{5}-1\right)+1=-\sqrt{5}+1+1=2-\sqrt{5}\)

1.

\(Q=\left[\frac{\sqrt{x}+2}{(\sqrt{x}+1)^2}-\frac{\sqrt{x}-2}{(\sqrt{x}-1)(\sqrt{x}+1)}\right].\sqrt{x}(\sqrt{x}+1)\)

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

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

2.

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

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

3.

\(A=\left[\frac{\sqrt{x}(\sqrt{x}-3)+2\sqrt{x}(\sqrt{x}+3)}{(\sqrt{x}-3)(\sqrt{x}+3)}-\frac{3x+9}{(\sqrt{x}-3)(\sqrt{x}+3)}\right]:\frac{1}{(\sqrt{x}+3)^2}\)

\(=\frac{3\sqrt{x}-9}{(\sqrt{x}-3)(\sqrt{x}+3)}.(\sqrt{x}+3)^2=\frac{3(\sqrt{x}-3)}{(\sqrt{x}-3)(\sqrt{x}+3)}(\sqrt{x}+3)^2=3(\sqrt{x}+3)\)