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

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

b: Để P nguyên thì \(\sqrt{x}+1⋮\sqrt{x}-1\)

\(\Leftrightarrow\sqrt{x}-1\in\left\{-1;1;2\right\}\)

hay \(x\in\left\{0;4;9\right\}\)

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

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

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

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

b: |P|>P

=>P<0

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

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

=>\(0<=x<1\)

c: \(P<\frac12\)

=>\(P-\frac12<0\)

=>\(\frac{\sqrt{x}-1}{\sqrt{x}+2}-\frac12<0\)

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

=>\(\sqrt{x}-4<0\)

=>\(\sqrt{x}<4\)

=>0<=x<16

Kết hợp ĐKXĐ, ta được: 0<=x<16 và x∉{1;4}

`C=(sqrtx+3)/(sqrtx-2)=(sqrtx-2+5)/(sqrtx-2)=1+5/(sqrtx-2)`

Ta cần tìm `max(5/(sqrtx-2))`

Nếu `0<=x<4` thì `5/(sqrtx-2)<0`

Nếu `x>4` thì `5/(sqrtx-2)>0`

Do đó ta chỉ xét `x>4` hay `x>=5(` Do `x` nguyên `)`

`=>sqrtx-2>=sqrt5-2`

`=>5/(sqrtx-2)<=5/(sqrt5-2)`

`=>C<=1+5/(sqrt5-2)=11+sqrt5`

Vậy `C_(max)=11+sqrt5<=>x=5`

a) \(P=\left(\dfrac{\sqrt{x}}{\sqrt{x}+1}+\dfrac{\sqrt{x}-4}{x-1}\right)\cdot\dfrac{\sqrt{x}+1}{\sqrt{x}-2}\left(dkxd:x\ge0;x\ne1;x\ne4\right)\)

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

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

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

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

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

b) Với \(x\ge0;x\ne1;x\ne4\):

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

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

\(=\dfrac{\sqrt{\left(\sqrt{2}\right)^2+2\cdot\sqrt{2}\cdot1+1^2}+2}{\sqrt{\left(\sqrt{2}\right)^2+2\cdot\sqrt{2}\cdot1+1^2}-1}\)

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

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

\(=\dfrac{\sqrt{2}+3}{\sqrt{2}}\)

\(=\dfrac{2+3\sqrt{2}}{2}\)

c) Với \(x\ge0;x\ne1;x\ne4\),

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

Để \(P\) có giá trị nguyên thì \(\dfrac{3}{\sqrt{x}-1}\) có giá trị nguyên

\(\Rightarrow 3\vdots\sqrt x-1\\\Rightarrow \sqrt x-1\in Ư(3)\)

\(\Rightarrow\sqrt{x}-1\in\left\{1;3;-1;-3\right\}\)

\(\Rightarrow\sqrt{x}\in\left\{2;4;0;-2\right\}\) mà \(\sqrt{x}\ge0\)

\(\Rightarrow\sqrt{x}\in\left\{2;4;0\right\}\)

\(\Rightarrow x\in\left\{4;16;0\right\}\)

Kết hợp với ĐKXĐ của \(x\), ta được:

\(x\in\left\{0;16\right\}\)

Vậy: ...

\(\text{#}Toru\)

a: Ta có: \(A=\frac{-x}{4-x}+\frac{1}{\sqrt{x}-2}+\frac{1}{\sqrt{x}+2}\)

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

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

b: Thay x=36 vào A, ta được:

\(A=\frac{6}{6-2}=\frac64=\frac32\)

c: \(A=-\frac13\)

=>\(\frac{\sqrt{x}}{\sqrt{x}-2}=-\frac13\)

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

=>\(4\sqrt{x}=2\)

=>\(\sqrt{x}=\frac12\)

=>\(x=\frac14\left(nhận\right)\)

e: A:B=-2

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

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

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

=>\(3\sqrt{x}=-2\) (vô lý)

=>x∈∅