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

Ta có: \(A=\dfrac{x-4\sqrt{x}+3-\left(2x-4\sqrt{x}-\sqrt{x}+2\right)+x+2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-1\right)}\)

\(=\dfrac{2x-4\sqrt{x}+5-2x+5\sqrt{x}-2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-1\right)}\)

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

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

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

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

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

\(=\dfrac{x-4\sqrt{x}+3-2x+5\sqrt{x}-2+x-2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-1\right)}\)
\(=\dfrac{\sqrt{x}-1}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-1\right)}=\dfrac{1}{\sqrt{x}-2}\)

b: Để A>2 thì A-2>0

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

=>\(\dfrac{5-2\sqrt{x}}{\sqrt{x}-2}>0\)

=>\(\dfrac{2\sqrt{x}-5}{\sqrt{x}-2}< 0\)

TH1: \(\left\{{}\begin{matrix}2\sqrt{x}-5>0\\\sqrt{x}-2< 0\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}\sqrt{x}>\dfrac{5}{2}\\\sqrt{x}< 2\end{matrix}\right.\)

=>\(x\in\varnothing\)

TH2: \(\left\{{}\begin{matrix}2\sqrt{x}-5< 0\\\sqrt{x}-2>0\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}\sqrt{x}< \dfrac{5}{2}\\\sqrt{x}>2\end{matrix}\right.\)

=>\(2< \sqrt{x}< \dfrac{5}{2}\)

=>4<x<25/4

c: Để A là số nguyên thì \(1⋮\sqrt{x}-2\)

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

=>\(\sqrt{x}\in\left\{3;1\right\}\)

=>\(x\in\left\{1;9\right\}\)

kết hợp ĐKXĐ, ta được: x=9

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

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

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

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

b) ta có : Q=3 => \(\dfrac{-2\sqrt{x}+1}{\sqrt{x}-1}=3=>-2\sqrt{x}+1=3\sqrt{x}-3\)

=>x=16/25=0,64

vậy x=0,64 khi Q=3

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

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

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

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

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

b) ta có : Q=3 <=> \(\sqrt{x}-1=3\)

\(\Leftrightarrow\) \(\sqrt{x}=4\Leftrightarrow x=16\)

vậy để Q=3 thì x=16

cái này ms đúng

a: Ta có: \(A=\dfrac{\sqrt{x}}{\sqrt{x}-1}+\dfrac{3}{\sqrt{x}+1}-\dfrac{6\sqrt{x}-4}{x-1}-1\)

\(=\dfrac{x+\sqrt{x}+3\sqrt{x}-4-6\sqrt{x}+4}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}-1\)

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

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

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

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

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

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

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

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

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

Để A là số nguyên thì \(2⋮x-1\)

\(\Leftrightarrow x-1\inƯ\left(2\right)\)

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

\(\Leftrightarrow x\in\left\{2;0;3;-1\right\}\)

Kết hợp ĐKXĐ, ta được: \(x\in\left\{2;3\right\}\)

Vậy: Để A là số nguyên thì \(x\in\left\{2;3\right\}\)

em tham khảo