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

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

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

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

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

b) Để \(P=5\) thì \(\dfrac{x+\sqrt{x}+1}{\sqrt{x}-1}=5\)

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

\(\Leftrightarrow x+\sqrt{x}+1=5\sqrt{x}-5\)

\(\Leftrightarrow x+\sqrt{x}+1-5\sqrt{x}+5=0\)

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

\(\Leftrightarrow x-2\cdot\sqrt{x}\cdot2+4+2=0\)

\(\Leftrightarrow\left(\sqrt{x}-2\right)^2+2=0\)(Vô lý)

Vậy: Không có giá trị nào của x để P=5

Lời giải:
ĐK: $x\geq 0; x\neq 4; x\neq 9$

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

\(=\frac{1}{\sqrt{x}+1}:\frac{x-9-(x-4)+\sqrt{x}+2}{(\sqrt{x}-2)(\sqrt{x}-3)}=\frac{1}{\sqrt{x}+1}:\frac{\sqrt{x}-3}{(\sqrt{x}-2)(\sqrt{x}-3)}=\frac{\sqrt{x}-2}{\sqrt{x}+1}\)

Để $P>0\Leftrightarrow \frac{\sqrt{x}-2}{\sqrt{x}+1}>0$

$\Leftrightarrow \sqrt{x}-2>0$ (do $\sqrt{x}+1>0$)

$\Leftrightarrow x>4$

Kết hợp với ĐKXĐ suy ra $x>4; x\neq 9$

a, \(P=\left(1-\dfrac{\sqrt{x}}{\sqrt{x}+1}\right):\left(\dfrac{\sqrt{x}+3}{\sqrt{x}-2}+\dfrac{\sqrt{x}+2}{3-\sqrt{x}}+\dfrac{\sqrt{x}+2}{x-5\sqrt{x}+6}\right)\)

\(P=\left(\dfrac{\sqrt{x}+1-\sqrt{x}}{\sqrt{x}+1}\right):\left(\dfrac{\sqrt{x}+3}{\sqrt{x}-2}-\dfrac{\sqrt{x}+2}{\sqrt{x}-3}+\dfrac{\sqrt{x}+2}{x-5\sqrt{x}+6}\right)\)

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

\(P=\dfrac{1}{\sqrt{x}+1}:\left[\dfrac{x-9-x+4+\sqrt{x}+2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}\right]\)

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

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

ĐKXĐ: \(\left\{{}\begin{matrix}x>0\\x\ne1\end{matrix}\right.\)

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

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

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

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

\(=2-x\)

b) Để P=3 thì 2-x=3

hay x=-1(Không thỏa mãn ĐKXĐ)

Vậy: Không có giá trị nào của x để P=3

c) Thay \(x=7+2\sqrt{3}\) vào P, ta được:

\(P=2-7-2\sqrt{3}=-5-2\sqrt{3}\)

Vậy: Khi \(x=7+2\sqrt{3}\) thì \(P=-5-2\sqrt{3}\)

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

a) Ta có: \(P=\dfrac{\sqrt{a}-1}{3\sqrt{a}+\left(\sqrt{a}-1\right)^2}-\dfrac{6-2\left(\sqrt{a}-1\right)^2}{a\sqrt{a}-1}+\dfrac{2}{\sqrt{a}-1}\)

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

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

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

\(=\dfrac{5a-4\sqrt{a}-1}{\left(\sqrt{a}-1\right)\left(a+\sqrt{a}+1\right)}\)

\(=\dfrac{5a-5\sqrt{a}+\sqrt{a}-1}{\left(\sqrt{a}-1\right)\left(a+\sqrt{a}+1\right)}\)

\(=\dfrac{5\sqrt{a}\left(\sqrt{a}-1\right)+\left(\sqrt{a}-1\right)}{\left(\sqrt{a}-1\right)\left(a+\sqrt{a}+1\right)}\)

\(=\dfrac{\left(\sqrt{a}-1\right)\left(5\sqrt{a}+1\right)}{\left(\sqrt{a}-1\right)\left(a+\sqrt{a}+1\right)}\)

\(=\dfrac{5\sqrt{a}+1}{a+\sqrt{a}+1}\)

b) Để P=1 thì \(5\sqrt{a}+1=a+\sqrt{a}+1\)

\(\Leftrightarrow a+\sqrt{a}+1-5\sqrt{a}-1=0\)

\(\Leftrightarrow a-4\sqrt{a}=0\)

\(\Leftrightarrow\sqrt{a}\left(\sqrt{a}-4\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}\sqrt{a}=0\\\sqrt{a}-4=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}a=0\left(nhận\right)\\a=16\left(nhận\right)\end{matrix}\right.\)

Vậy: Để P=1 thì \(a\in\left\{0;16\right\}\)

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

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

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

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

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

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

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:<

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

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

2: P<1/2
=>P-1/2<0

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

=>-x+2căn x-1<0

=>(căn x-1)^2>0(luôn đúng)

a: ĐKXĐ: x>1; x<>2

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

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

c: Khi x=3+2căn 2 thì

P=(-căn 2-1+căn 2)/(căn 2+1)=căn 2-1