Bài 1: 

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

b) \(\dfrac{\sqrt{\left(x-3\right)^2}}{3-x}=\dfrac{\left|x-3\right|}{3-x}=\pm1\)

Bài 2: 

a) \(\dfrac{\sqrt{9x^2-6x+1}}{9x^2-1}=\dfrac{\left|3x-1\right|}{\left(3x-1\right)\left(3x+1\right)}=\pm\dfrac{1}{3x+1}\)

b) \(4-x-\sqrt{x^2-4x+4}=4-x-\left|x-2\right|=\left[{}\begin{matrix}6-2x\left(x\ge2\right)\\2\left(x< 2\right)\end{matrix}\right.\)

a: Ta có: \(3\sqrt{5a}-\sqrt{20a}+\sqrt{45a}\)

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

\(=4\sqrt{5a}\)

b: Ta có: \(\sqrt{160a^2}+\dfrac{1}{2}\sqrt{40a^2}-3\sqrt{90a^2}\)

\(=4a\sqrt{10}+\dfrac{1}{2}\cdot2a\sqrt{10}-3\cdot3a\sqrt{10}\)

\(=-4a\sqrt{10}\)

c: Ta có: \(\sqrt{x^2-2x+1}-\sqrt{x^2-4x+4}\)

\(=\left|x-1\right|-\left|x-2\right|\)

A. ĐKXĐ: $x>0; x\neq 1; x\neq 4$

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

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

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

B.

ĐKXĐ: $x\geq 0, x\neq \frac{1}{4}$

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

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

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

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

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

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

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

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

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

Ta có: \(\dfrac{8+x\left(1+\sqrt{x-2\sqrt{x}+1}\right)}{\left(x-4\right)\left(x-2\sqrt{x}+4\right)}+\dfrac{x-3\sqrt{x}}{2\left(x-\sqrt{x}-6\right)}\)

\(=\dfrac{8+x\left(1+\sqrt{x}-1\right)}{\left(x-4\right)\left(x-2\sqrt{x}+4\right)}+\dfrac{\sqrt{x}\left(\sqrt{x}-3\right)}{2\left(\sqrt{x}-3\right)\left(\sqrt{x}+2\right)}\)

\(=\dfrac{x\sqrt{x}+8}{\left(x-4\right)\left(x-2\sqrt{x}+4\right)}+\dfrac{\sqrt{x}}{2\left(\sqrt{x}+2\right)}\)

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

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

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

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

\(=\dfrac{x+4}{2x-8}\)

\(a,=\dfrac{x}{y}\cdot\dfrac{\left|x\right|}{y^2}=\dfrac{x^2}{y^3}\\ b,=2y^2\cdot\dfrac{x^2}{\left|2y\right|}=\dfrac{2x^2y^2}{-2y}=-x^2y\)

1:

a: ĐKXĐ: 1-x>=0

=>x<=1

b: ĐKXĐ: 2/x>=0

=>x>0

c: ĐKXĐ: 4/x+1>=0

=>x+1>0

=>x>-1

d: ĐKXĐ: x^2+2>=0

=>x thuộc R

Câu 2:

a: \(=\left|-\sqrt{2-1}\right|=\sqrt{1}=1\)

b: \(=\left|4+\sqrt{2}\right|=4+\sqrt{2}\)

\(A=\left(\dfrac{2\sqrt{x}}{\sqrt{x}-3}+\dfrac{x}{3\sqrt{x}-x}\right).\dfrac{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}{\sqrt{x}+3}\left(dkxd:x\ne0;\pm\sqrt{3}\right)\)

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

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

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

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

Vậy \(A=2-\sqrt{x}\)

\(1,A=\dfrac{2x+1-x}{\left(\sqrt{x}+1\right)\left(x-\sqrt{x}+1\right)}\cdot\left(x-\sqrt{x}-2\right)\\ A=\dfrac{\left(x+1\right)\left(\sqrt{x}+1\right)\left(\sqrt{x}-2\right)}{\left(\sqrt{x}+1\right)\left(x-\sqrt{x}+1\right)}=\dfrac{\left(x+1\right)\left(\sqrt{x}-2\right)}{x-\sqrt{x}+1}\\ 2,\Leftrightarrow\left\{{}\begin{matrix}2a-b=1\\a-b=2\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}a=-1\\b=-3\end{matrix}\right.\Leftrightarrow y=-x-3\)