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

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

b: \(M=\left(\dfrac{\sqrt{a}+a}{\sqrt{a}-2}\right)\cdot\dfrac{\left(\sqrt{a}-2\right)^2}{\sqrt{a}+1}\)

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

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

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

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

\(M=\left(\dfrac{a}{a-2\sqrt{a}}+\dfrac{a}{\sqrt{a}-2}\right):\dfrac{\sqrt{a}+1}{a-4\sqrt{a}+4}\)

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

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

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

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

\(\dfrac{\sqrt{a}+\sqrt{b}}{2\sqrt{a}-2\sqrt{b}}-\dfrac{\sqrt{a}-\sqrt{b}}{2\sqrt{a}+2\sqrt{b}}-\dfrac{2b}{b-a}\left(a,b>0;a\ne b\right)\\ =\dfrac{\left(\sqrt{a}+\sqrt{b}\right)^2-\left(\sqrt{a}-\sqrt{b}\right)^2+4b}{2\left(\sqrt{a}-\sqrt{b}\right)\left(\sqrt{a}+\sqrt{b}\right)}\\ =\dfrac{4\sqrt{ab}+4b}{2\left(\sqrt{a}-\sqrt{b}\right)\left(\sqrt{a}+\sqrt{b}\right)}\\ =\dfrac{4\sqrt{b}\left(\sqrt{a}+\sqrt{b}\right)}{2\left(\sqrt{a}-\sqrt{b}\right)\left(\sqrt{a}+\sqrt{b}\right)}=\dfrac{2\sqrt{b}}{\sqrt{a}-\sqrt{b}}\)

Tick plz

Ta có: \(\dfrac{\sqrt{a}+\sqrt{b}}{2\sqrt{a}-2\sqrt{b}}-\dfrac{\sqrt{a}-\sqrt{b}}{2\sqrt{a}+2\sqrt{b}}-\dfrac{2b}{b-a}\)

\(=\dfrac{a+2\sqrt{ab}+b-a+2\sqrt{ab}-b+4b}{2\left(\sqrt{a}-\sqrt{b}\right)\left(\sqrt{a}+\sqrt{b}\right)}\)

\(=\dfrac{4b+4\sqrt{ab}}{2\left(\sqrt{a}-\sqrt{b}\right)\left(\sqrt{a}+\sqrt{b}\right)}\)

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

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

Câu 2: 

a: \(=2\left(\sqrt{4+\sqrt{5}-1}\right)\left(\sqrt{10}-\sqrt{2}\right)\)

\(=\sqrt{2}\cdot\sqrt{6+2\sqrt{5}}\cdot\left(\sqrt{10}-\sqrt{2}\right)\)

\(=2\cdot\left(\sqrt{5}+1\right)\left(\sqrt{5}-1\right)=8\)

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

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

1) \(VT=\frac{\sqrt{a}+\sqrt{b}}{2\left(\sqrt{a}-\sqrt{b}\right)}-\frac{\sqrt{a}-\sqrt{b}}{2\left(\sqrt{a}+\sqrt{b}\right)}+\frac{2b}{\left(\sqrt{a}+\sqrt{b}\right)\left(\sqrt{a}-\sqrt{b}\right)}\)

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

\(=\frac{4\sqrt{ab}+4b}{2\left(\sqrt{a}+\sqrt{b}\right)\left(\sqrt{a}-\sqrt{b}\right)}\)

\(=\frac{4\sqrt{b}\left(\sqrt{a}+\sqrt{b}\right)}{2\left(\sqrt{a}+\sqrt{b}\right)\left(\sqrt{a}-\sqrt{b}\right)}=\frac{2\sqrt{b}}{\sqrt{a}-\sqrt{b}}=VP\)(ĐPCM)

2) \(VT=\text{[}\frac{\left(\sqrt{a}+\sqrt{b}\right)\left(a+b-\sqrt{ab}\right)}{\left(\sqrt{a}+\sqrt{b}\right)}-\sqrt{ab}\text{]}.\frac{\left(\sqrt{a}+\sqrt{b}\right)^2}{\left(a-b\right)^2}\)

\(=\frac{\left(a+b-\sqrt{ab}-\sqrt{ab}\right)\left(\sqrt{a}+\sqrt{b}\right)^2}{\left(a-b\right)^2}\)\(=\frac{\left(\sqrt{a}-\sqrt{b}\right)^2\left(\sqrt{a}+\sqrt{b}\right)^2}{\left(a-b\right)^2}=\frac{\left(a-b\right)^2}{\left(a-b\right)^2}=1=VP\)(ĐPCM)

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

\(=\left(1+\sqrt{a}\right)\left(1-\sqrt{a}\right)=1-a=VP\)(ĐPCM)

a) Đặt \(x=\sqrt{a+\sqrt{b}}+\sqrt{a-\sqrt{b}}\)Vì x > 0 \(\Rightarrow x=\sqrt{x^2}\)

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

Tương tự : Đặt \(y=\sqrt{a+\sqrt{b}}-\sqrt{a-\sqrt{b}}\) 

Xét biểu thức \(\sqrt{a+\sqrt{b}}-\sqrt{a-\sqrt{b}}>0\Leftrightarrow a+\sqrt{b}>a-\sqrt{b}\Leftrightarrow\sqrt{b}>0\)(luôn đúng)

Do đó : \(y>0\) \(\Rightarrow y=\sqrt{y^2}\)

Ta có : \(y^2=2a-2\sqrt{a^2-b}=4\left(\frac{a-\sqrt{a^2-b}}{2}\right)\Rightarrow y=2\sqrt{\frac{a-\sqrt{a^2-b}}{2}}\)hay \(\sqrt{a+\sqrt{b}}-\sqrt{a-\sqrt{b}}=2\sqrt{\frac{a-\sqrt{a^2-b}}{2}}\)(2)

Cộng (1) và (2) theo vế ta được : \(\sqrt{a+\sqrt{b}}=\sqrt{\frac{a+\sqrt{a^2-b}}{2}}+\sqrt{\frac{a-\sqrt{a^2-b}}{2}}\)(đpcm)

Câu b) bạn làm tương tự nhé!

Căn thức phức tạp trên mạng có

hộ ttis ạ

ĐKXĐ: ...

\(A=\left(\frac{\sqrt{a}+2}{\left(\sqrt{a}+1\right)^2}-\frac{\sqrt{a}-2}{\left(\sqrt{a}+1\right)\left(\sqrt{a}-1\right)}\right)\frac{\sqrt{a}+1}{\sqrt{a}}\)

\(=\left(\frac{\sqrt{a}+2}{\sqrt{a}+1}-\frac{\sqrt{a}-2}{\sqrt{a}-1}\right).\frac{\left(\sqrt{a}+1\right)}{\sqrt{a}\left(\sqrt{a}+1\right)}=\left(\frac{\left(\sqrt{a}+2\right)\left(\sqrt{a}-1\right)-\left(\sqrt{a}-2\right)\left(\sqrt{a}+1\right)}{\left(\sqrt{a}+1\right)\left(\sqrt{a}-1\right)}\right)\frac{1}{\sqrt{a}}\)

\(=\left(\frac{a+\sqrt{a}-2-a+\sqrt{a}+2}{\left(\sqrt{a}-1\right)\left(\sqrt{a}+1\right)}\right).\frac{1}{\sqrt{a}}=\frac{2}{a-1}\)

\(B=\sqrt{\frac{a+\sqrt{a^2-b}}{2}}+\sqrt{\frac{a-\sqrt{a^2-b}}{2}}\) \(\Rightarrow B\ge0\)

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

\(\Rightarrow B=\sqrt{a+\sqrt{b}}\)