1:

\(A=\sqrt{x^2+\dfrac{2x^2}{3}}=\sqrt{\dfrac{5x^2}{3}}=\left|\sqrt{\dfrac{5}{3}}x\right|=-x\sqrt{\dfrac{5}{3}}\)

2: \(=\left(\dfrac{\sqrt{100}+\sqrt{40}}{\sqrt{5}+\sqrt{2}}+\sqrt{6}\right)\cdot\dfrac{2\sqrt{5}-\sqrt{6}}{2}\)

\(=\dfrac{\left(2\sqrt{5}+\sqrt{6}\right)\left(2\sqrt{5}-\sqrt{6}\right)}{2}\)

\(=\dfrac{20-6}{2}=7\)

\(A=\dfrac{2\sqrt{3}}{\sqrt{3}+1}+3\sqrt{\dfrac{1}{6}}\cdot\sqrt{\dfrac{1}{2}}-\sqrt{12}\)

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

\(A=\dfrac{2\sqrt{3}\cdot\left(\sqrt{3}-1\right)}{2}+3\cdot\dfrac{1}{\sqrt{12}}-2\sqrt{3}\)

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

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

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

\(A=\dfrac{6+6\sqrt{3}+\sqrt{3}}{2}\)

\(A=\dfrac{6+7\sqrt{3}}{2}\)

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

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

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

\(=\sqrt{\dfrac{4\sqrt{3}}{2}}=\sqrt{2\sqrt{3}}\)

Lời giải:

a) ĐK: $x\geq 0; y\geq 0; x\neq y$

\(A=\left[\frac{(\sqrt{x}-\sqrt{y})(\sqrt{x}+\sqrt{y})}{\sqrt{x}-\sqrt{y}}-\frac{(\sqrt{x}-\sqrt{y})(x+\sqrt{xy}+y)}{(\sqrt{x}-\sqrt{y})(\sqrt{x}+\sqrt{y})}\right]:\frac{x-\sqrt{xy}+y}{\sqrt{x}+\sqrt{y}}\)

\(=\left(\sqrt{x}+\sqrt{y}-\frac{x+\sqrt{xy}+y}{\sqrt{x}+\sqrt{y}}\right).\frac{\sqrt{x}+\sqrt{y}}{x-\sqrt{xy}+y}\)

\(=\frac{\sqrt{xy}}{\sqrt{x}+\sqrt{y}}.\frac{\sqrt{x}+\sqrt{y}}{x-\sqrt{xy}+y}=\frac{\sqrt{xy}}{x-\sqrt{xy}+y}\)

b) \(1-A=\frac{(\sqrt{x}-\sqrt{y})^2}{x-\sqrt{xy}+y}>0\) với mọi $x\neq y; x,y\geq 0$

$\Rightarrow A< 1$

Lời giải:

$\sqrt{7+2\sqrt{10}}=\sqrt{2+5+2\sqrt{2.5}}=\sqrt{(\sqrt{2}+\sqrt{5})^2}=\sqrt{2}+\sqrt{5}$

\(\sqrt[3]{3\sqrt[3]{3}-3\sqrt[3]{2}-1}=\sqrt[3]{(1-\sqrt[3]{2})^3}=1-\sqrt[3]{2}\)

Do đó:

\(\text{TS}=\sqrt[3]{2}+\sqrt{2}+\sqrt{5}+1-\sqrt[3]{2}=\sqrt{2}+\sqrt{5}+1=\text{MS}\)

\(A=\frac{\text{TS}}{\text{MS}}=1\)

\(A=\dfrac{1}{2-\sqrt{3}}+\dfrac{1}{2+\sqrt{3}}-\sqrt{37-20\sqrt{3}}\)

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

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

\(=2\sqrt{3}-1\)

`A=(1/(2-sqrt3)-1/(2+sqrt3)).(sqrt3-1)/(3-sqrt3)`

`=((2+sqrt3)/(4-3)-(2-sqrt3)/(4-3)).((sqrt3-1)/(sqrt3(sqrt3-1)))`

`=(2+sqrt3-2+sqrt3).1/(sqrt3)`

`=2sqrt3. 1/sqrt3`

`=2`

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

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

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

a)Sửa đề:\(\sqrt{3}+\sqrt{8-2\sqrt{15}}=\sqrt{3}+\sqrt{5-2\sqrt{5}\sqrt{3}+3}=\sqrt{3}+\sqrt{\left(\sqrt{5}-\sqrt{3}\right)^2}\)

\(=\sqrt{3}+\left|\sqrt{5}-\sqrt{3}\right|=\sqrt{3}+\sqrt{5}-\sqrt{3}=\sqrt{5}\)

b)Đk: \(x\ge2\)

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

TH1:\(\sqrt{x-2}-1>0\Leftrightarrow x>3\)

\(\Rightarrow A=\sqrt{x-2}-1\)

TH2:\(\sqrt{x-2}-1< 0\Leftrightarrow2\le x< 3\)

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

Vậy...

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