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

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

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

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

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

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

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

a, \(A=\left(\dfrac{2\sqrt{x}}{\sqrt{x}+3}+\dfrac{\sqrt{x}}{\sqrt{x}-3}-\dfrac{3x+3}{x-9}\right):\left(\dfrac{2\sqrt{x}-2}{\sqrt{x}-3}-1\right)\)

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

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

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

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

b, \(A\in Z\Leftrightarrow\dfrac{-3}{\sqrt{x}+3}\in Z\)

\(\Leftrightarrow\sqrt{x}+3\inƯ_3=\left\{\pm1;\pm3\right\}\)

\(\Leftrightarrow\sqrt{x}=0\)

\(\Leftrightarrow x=0\)

\(a,A=\left(\dfrac{2\sqrt{x}}{\sqrt{x}+3}+\dfrac{\sqrt{x}}{\sqrt{x}-3}-\dfrac{3x+3}{x-9}\right):\left(\dfrac{2\sqrt{x}-2}{\sqrt{x}-3}-1\right)\left(x\ge0;x\ne9\right)\\ A=\dfrac{2\sqrt{x}\left(\sqrt{x}-3\right)+\sqrt{x}\left(\sqrt{x}+3\right)-3x-3}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}:\dfrac{2\sqrt{x}-2-\sqrt{x}+3}{\sqrt{x}-3}\\ A=\dfrac{2x-6\sqrt{x}+x+3\sqrt{x}-3x-3}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}\cdot\dfrac{\sqrt{x}-3}{\sqrt{x}+1}\\ A=\dfrac{-3\left(\sqrt{x}+1\right)}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}\cdot\dfrac{\sqrt{x}-3}{\sqrt{x}+1}=\dfrac{-3}{\sqrt{x}+3}\)

\(b,A\in Z\Leftrightarrow\dfrac{-3}{\sqrt{x}+3}\in Z\Leftrightarrow-3⋮\sqrt{x}+3\\ \Leftrightarrow\sqrt{x}+3\inƯ\left(-3\right)=\left\{-3;-1;1;3\right\}\\ \Leftrightarrow\sqrt{x}\in\left\{-6;-4;-2;0\right\}\)

Mà \(\sqrt{x}\ge0\)

\(\Leftrightarrow x\in\left\{0\right\}\)

Vậy \(x=0\) thì A nguyên

ĐKXĐ: \(x\ge0,x\ne1\)

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

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

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

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

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

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

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

b: Khi x=9/4 thì \(P=\dfrac{3}{2}:\left(\dfrac{3}{2}-1\right)=\dfrac{3}{2}:\dfrac{1}{2}=3\)

c: P<0

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

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

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

=>0<=x<1

ĐKXĐ: \(x\ge0,x\ne1\)

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

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

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

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

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

b) Để \(P=\dfrac{3}{2}\) thì \(4\sqrt{x}+2=3\sqrt{x}+3\)

\(\Leftrightarrow x=1\)(Vô lý)

Lời giải:

A có min thôi bạn nhé.

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

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

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

Vì $\sqrt{x}\geq 0$ với mọi $x\geq 0; x\neq 1$ nên $\sqrt{x}+1\geq 1$

$\Rightarrow \frac{1}{\sqrt{x}+1}\leq 1$
$\Rightarrow A=2-\frac{1}{\sqrt{x}+1}\geq 2-1=1$

Vậy $A_{\min}=1$ tại $x=0$

a, x > 0 ; x khác 1 

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

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

b, Ta có : \(P=\dfrac{x-2}{\sqrt{x}}=1\Rightarrow x-2=\sqrt{x}\)

\(\Leftrightarrow x-\sqrt{x}-2=0\Leftrightarrow\left(\sqrt{x}+1>0\right)\left(\sqrt{x}-2\right)=0\Leftrightarrow x=4\)(tm) 

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

b: Để P=1 thì \(x-\sqrt{x}-2=0\)

hay x=4

Câu a có sai đề nên mk có sửa lại nha ​