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

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

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

=\(\dfrac{a-b}{a-b}=1=VP\)

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

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

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

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

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

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

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

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

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

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

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

Tick plz

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

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

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

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

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

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

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

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

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

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

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

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

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

a) Ta có:

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

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

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

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

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

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

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

(với a, b không âm và a ≠b )

Vế trái bằng vế phải nên đẳng thức được chứng minh.

b) Ta có:

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

\(=\left(\dfrac{\sqrt{a^3}+\sqrt{b^3}}{\sqrt{a}+\sqrt{b}}-\sqrt{ab}\right)\left[\dfrac{\sqrt{a}+\sqrt{b}}{\left(\sqrt{a}+\sqrt{b}\right)\left(\sqrt{a}-\sqrt{b}\right)}\right]^2\)

\(=\left[\dfrac{\left(\sqrt{a}+\sqrt{b}\right)\left(\sqrt{a^2}-\sqrt{ab}+\sqrt{b}\right)}{\sqrt{a}+\sqrt{b}}-\sqrt{ab}\right]\left(\dfrac{1}{\sqrt{a}-\sqrt{b}}\right)^2\)

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

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

(với a, b không âm và a ≠b )

Vế trái bằng vế phải nên đẳng thức được chứng minh.

ai giúp vs

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

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

\(\Leftrightarrow2-\sqrt{3}+2+\sqrt{3}=4\)

\(\Leftrightarrow4=4\left(đpcm\right)\)

2. \(\dfrac{\sqrt{a}}{\sqrt{a}-\sqrt{b}}-\dfrac{\sqrt{b}}{\sqrt{a}+\sqrt{b}}-\dfrac{2b}{a-b}=1\)

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

\(\Leftrightarrow\dfrac{a+\sqrt{ab}-\sqrt{ab}+b-2b}{a-b}1\)

\(\Leftrightarrow\dfrac{a-b}{a-b}=1\left(đpcm\right)\)