\(x=\dfrac{1}{2}\cdot\left(\dfrac{a}{\sqrt{ab}}+\dfrac{b}{\sqrt{ab}}\right)=\dfrac{a+b}{2\sqrt{ab}}\)

\(2\sqrt{x^2}-1=2\cdot\dfrac{a+b}{2\sqrt{ab}}-1=\dfrac{a+b-\sqrt{ab}}{\sqrt{ab}}\)

\(x-\sqrt{x^2-1}=\dfrac{a+b}{2\sqrt{ab}}-\sqrt{\dfrac{a^2+2ab+b^2}{4ab}-1}\)

\(=\dfrac{a+b}{2\sqrt{ab}}-\dfrac{a-b}{2\sqrt{ab}}=\dfrac{2b}{2\sqrt{ab}}=\dfrac{\sqrt{b}}{\sqrt{a}}\)

\(G=\dfrac{a+b-\sqrt{ab}}{\sqrt{ab}}:\dfrac{\sqrt{b}}{\sqrt{a}}=\dfrac{a+b-\sqrt{ab}}{\sqrt{ab}}\cdot\dfrac{\sqrt{a}}{\sqrt{b}}\)

\(=\dfrac{a+b-\sqrt{ab}}{b}\)

A)

Đặt \(\sqrt{1+2x}=a; \sqrt{1-2x}=b\) (\(a,b>0\) )

\(\Rightarrow \left\{\begin{matrix} a^2+b^2=2\\ a^2-b^2=4x=\sqrt{3}\end{matrix}\right.\)

\(\Rightarrow \left\{\begin{matrix} 2a^2=2+\sqrt{3}\rightarrow 4a^2=4+2\sqrt{3}=(\sqrt{3}+1)^2\\ 2b^2=2-\sqrt{3}\rightarrow 4b^2=4-2\sqrt{3}=(\sqrt{3}-1)^2\end{matrix}\right.\)

\(\Rightarrow a=\frac{\sqrt{3}+1}{2}; b=\frac{\sqrt{3}-1}{2}\)

\(\Rightarrow ab=\frac{(\sqrt{3}+1)(\sqrt{3}-1)}{4}=\frac{1}{2}; a-b=1\)

Có:

\(A=\frac{a^2}{1+a}+\frac{b^2}{1-b}=\frac{a^2-a^2b+b^2+ab^2}{(1+a)(1-b)}\)

\(=\frac{2-ab(a-b)}{1+(a-b)-ab}=\frac{2-\frac{1}{2}.1}{1+1-\frac{1}{2}}=1\)

B)

\(2x=\sqrt{\frac{a}{b}}+\sqrt{\frac{b}{a}}\)

\(\Rightarrow 4x^2=\frac{a}{b}+\frac{b}{a}+2\)

\(\rightarrow 4(x^2-1)=\frac{a}{b}+\frac{b}{a}-2=\left(\sqrt{\frac{a}{b}}-\sqrt{\frac{b}{a}}\right)^2\)

\(\Rightarrow \sqrt{4(x^2-1)}=\sqrt{\frac{a}{b}}-\sqrt{\frac{b}{a}}\) do $a>b$

T có: \(B=\frac{b\sqrt{4(x^2-1)}}{x-\sqrt{x^2-1}}=\frac{2b\sqrt{4(x^2-1)}}{2x-\sqrt{4(x^2-1)}}=\frac{2b\left ( \sqrt{\frac{a}{b}}-\sqrt{\frac{b}{a}} \right )}{\sqrt{\frac{a}{b}}+\sqrt{\frac{b}{a}}-\left ( \sqrt{\frac{a}{b}}-\sqrt{\frac{b}{a}} \right )}\)

\(=\frac{2b\left ( \sqrt{\frac{a}{b}}-\sqrt{\frac{b}{a}} \right )}{2\sqrt{\frac{b}{a}}}=\frac{b\left ( \sqrt{\frac{a}{b}}-\sqrt{\frac{b}{a}} \right )}{\sqrt{\frac{b}{a}}}=\frac{\frac{b(a-b)}{\sqrt{ab}}}{\sqrt{\frac{b}{a}}}=a-b\)

Bài 2: 

a: \(=\sqrt{\left(\dfrac{1}{5a}\right)^2}=\dfrac{1}{\left|5a\right|}=\dfrac{-1}{5a}\)

b: \(=\dfrac{1}{3}\cdot15\cdot\left|a\right|=5\left|a\right|\)

a, Vì trong dấu căn là số âm nên biểu thức này vô nghĩa. b)\(\sqrt{\dfrac{1}{200}}=\dfrac{1}{\sqrt{200}}=\dfrac{1}{10\sqrt{2}}=\dfrac{\sqrt{2}}{10\sqrt{2}.\sqrt{2}}=\dfrac{\sqrt{2}}{20}\)

c,\(\sqrt{\dfrac{7}{500}}=\dfrac{\sqrt{7}}{\sqrt{500}}=\dfrac{\sqrt{7}}{10\sqrt{5}}=\dfrac{\sqrt{7}.\sqrt{5}}{10\sqrt{5}.\sqrt{5}}=\dfrac{\sqrt{35}}{50}\)

a: \(=2ab\cdot\dfrac{-15}{b^2a}=\dfrac{-30}{b}\)

b: \(=\dfrac{2}{3}\cdot\left(1-a\right)=\dfrac{2}{3}-\dfrac{2}{3}a\)

c: \(=\dfrac{\left|3a-1\right|}{\left|b\right|}=\dfrac{3a-1}{b}\)

d: \(=\left(a-2\right)\cdot\dfrac{a}{-\left(a-2\right)}=-a\)

ta có cos60=1/2

sin 60=\(\frac{\sqrt{3}}{2}\)

tan 30=\(\frac{\sqrt{3}}{3}\)

ta thay vào biểu thức trên

=> \(\frac{\frac{1}{2}}{1+\frac{\sqrt{3}}{2}}+\frac{1}{\frac{\sqrt{3}}{3}}=2\)

\(\frac{cos60^o}{1+sin60^o}+\frac{1}{tan30^o}=\frac{\frac{1}{2}}{1+\frac{\sqrt{3}}{2}}+\frac{1}{\frac{\sqrt{3}}{3}}=\frac{1}{2}.\frac{2}{\sqrt{3}+2}+\sqrt{3}=\frac{1}{\sqrt{3}+2}+\sqrt{3}\)

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