ĐKXĐ: x>=0; x<>4

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

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

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

ĐKXĐ: x>0; x<>4

\(\dfrac{3x-4}{x-2\sqrt{x}}-\dfrac{\sqrt{x}+2}{\sqrt{x}}+\dfrac{\sqrt{x}-1}{2-\sqrt{x}}\)

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

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

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

ĐKXĐ: a>=0; a<>9

\(Q=\dfrac{3}{\sqrt{a}-3}+\dfrac{2}{\sqrt{a}+3}-\dfrac{a-5\sqrt{a}-3}{9-a}\)

\(=\dfrac{3\left(\sqrt{a}+3\right)+2\left(\sqrt{a}-3\right)+a-5\sqrt{a}-3}{a-9}\)

\(=\dfrac{3\sqrt{a}+9+2\sqrt{a}-6+a-5\sqrt{a}-3}{a-9}=\dfrac{a}{a-9}\)

a: 

ĐKXĐ: x>0; x<>1\(M=\left(\dfrac{\sqrt{x}}{\sqrt{x}-1}-\dfrac{3\sqrt{x}-1}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\right):\dfrac{\left(\sqrt{x}+1\right)^2-4\sqrt{x}-1}{\sqrt{x}\left(\sqrt{x}+1\right)}\)

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

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

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

b: M là số nguyên

=>\(\sqrt{x}-1⋮\sqrt{x}-2\)

=>\(\sqrt{x}-2+1⋮\sqrt{x}-2\)

=>căn x-2 thuộc {1;-1}

=>căn x thuộc {3;1}

=>x thuộc {9;1}

Kết hợp ĐKXĐ, ta được: x=9

c: M<0

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

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

=>1<x<4

a) a ≠ 1; a ≥ 0

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

b) a ≥ 0; \(x\ne\pm\sqrt{3}\)

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

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

Ta có: \(\dfrac{a-5\sqrt{a}+4}{a-1}\)

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

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

2) ĐKXĐ: \(\left\{{}\begin{matrix}x\ge0\\x\ne\sqrt{3}\end{matrix}\right.\)

Ta có: \(\dfrac{\sqrt{x^2+2\sqrt{3x}+3}}{x^2-3}\)

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

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

Đk: \(x\ge4\)

\(A=\sqrt{x+4\sqrt{x-4}}+\sqrt{x-4\sqrt{x-4}}\)

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

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

\(=\sqrt{x-4}+2+\left|\sqrt{x-4}-2\right|\)

TH1:\(\sqrt{x-4}>2\Leftrightarrow x>8\)

\(A=\sqrt{x-4}+2+\sqrt{x-4}-2=2\sqrt{x-4}\)

TH2:\(\sqrt{x-4}\le2\Leftrightarrow4\le x\le8\)

\(A=\sqrt{x-4}+2-\left(\sqrt{x-4}-2\right)=4\)

Vậy...

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

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

a) ĐKXĐ: \(x\notin\left\{0;3;1\right\}\)

Sửa đề: \(A=\left(\dfrac{x-3}{x}-\dfrac{x}{x-3}+\dfrac{9}{x^2-3x}\right):\dfrac{2x-2}{x}\)

Ta có: \(A=\left(\dfrac{x-3}{x}-\dfrac{x}{x-3}+\dfrac{9}{x^2-3x}\right):\dfrac{2x-2}{x}\)

\(=\dfrac{x^2-6x+9-x^2+9}{x\left(x-3\right)}\cdot\dfrac{x}{2\left(x-1\right)}\)

\(=\dfrac{-6x+18}{x\left(x-3\right)}\cdot\dfrac{x}{2\left(x-1\right)}\)

\(=\dfrac{-6\left(x-3\right)}{x\left(x-3\right)}\cdot\dfrac{x}{2\left(x-1\right)}\)

\(=\dfrac{-3}{x-1}\)

b) Để A nguyên thì \(-3⋮x-1\)

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

\(\Leftrightarrow x\in\left\{2;0;4;-2\right\}\)

Kết hợp ĐKXĐ, ta được: \(x\in\left\{2;-2;4\right\}\)