f: \(\dfrac{2}{\sqrt{6}-\sqrt{5}}=2\sqrt{6}+2\sqrt{5}\)

l: \(\dfrac{3}{\sqrt{10}+\sqrt{7}}=\sqrt{10}-\sqrt{7}\)

a: \(\dfrac{5}{\sqrt{10}}=\dfrac{5\sqrt{10}}{10}=\dfrac{\sqrt{10}}{2}\)

b: \(\dfrac{5}{2\sqrt{5}}=\dfrac{\sqrt{5}}{2}\)

c: \(\dfrac{1}{3\sqrt{20}}=\dfrac{\sqrt{5}}{30}\)

a)\(\dfrac{5}{\sqrt{10}}=\dfrac{5\sqrt{10}}{10}=\dfrac{\sqrt{10}}{2}\)

b)\(\dfrac{5}{2\sqrt{5}}=\dfrac{5\sqrt{5}}{2.5}=\dfrac{\sqrt{5}}{2}\)

c)\(\dfrac{1}{3\sqrt{20}}=\dfrac{\sqrt{20}}{3.20}=\dfrac{\sqrt{20}}{60}=\dfrac{\sqrt{5}}{30}\)

d)\(\dfrac{2\sqrt{2}+2}{5\sqrt{2}}=\dfrac{2.2+2\sqrt{2}}{5.2}=\dfrac{2+\sqrt{2}}{5}\)

e)\(\dfrac{y+b\sqrt{y}}{b\sqrt{y}}=\dfrac{y\sqrt{y}+by}{by}=\dfrac{\sqrt{y}+b}{b}\)

a: \(=\dfrac{3}{2}\sqrt{6}+\dfrac{2}{3}\sqrt{6}-2\sqrt{3}=\dfrac{13}{6}\sqrt{6}-2\sqrt{3}\)

b: \(VT=\dfrac{\sqrt{xy}\left(\sqrt{x}+\sqrt{y}\right)}{\sqrt{xy}}\cdot\left(\sqrt{x}+\sqrt{y}\right)=\left(\sqrt{x}+\sqrt{y}\right)^2\)

c: \(VT=\dfrac{\sqrt{y}}{\sqrt{x}\left(\sqrt{x}-\sqrt{y}\right)}+\dfrac{\sqrt{x}}{\sqrt{y}\left(\sqrt{y}-\sqrt{x}\right)}\)

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

a: \(=\dfrac{3}{2}\sqrt{6}+\dfrac{2}{3}\sqrt{6}-2\sqrt{6}\)

\(=\dfrac{1}{6}\sqrt{6}\)

b: \(VT=\dfrac{\sqrt{y}}{\sqrt{x}\left(\sqrt{x}-\sqrt{y}\right)}+\dfrac{\sqrt{x}}{\sqrt{y}\left(\sqrt{y}-\sqrt{x}\right)}\)

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

Bài 1: 

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

\(=\dfrac{x+2\sqrt{xy}+y-x-\sqrt{xy}-y}{\sqrt{x}+\sqrt{y}}\cdot\dfrac{\sqrt{x}+\sqrt{y}}{x-\sqrt{xy}+y}\)

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

b: \(\sqrt{xy}>=0;x-\sqrt{xy}+y>0\)

Do đó: A>=0

1) ta có : \(P=\dfrac{x\sqrt{y}+y\sqrt{x}}{\sqrt{xy}}:\dfrac{1}{\sqrt{x}-\sqrt{y}}\)

\(\Leftrightarrow P=\dfrac{\sqrt{xy}\left(\sqrt{x}+\sqrt{y}\right)}{\sqrt{xy}}.\left(\sqrt{x}-\sqrt{y}\right)\)

\(\Leftrightarrow P=\left(\sqrt{x}+\sqrt{y}\right)\left(\sqrt{x}-\sqrt{y}\right)=x-y\)

2) ta có : \(B=\dfrac{\sqrt{2+\sqrt{3}}}{\sqrt{2}}:\left(\dfrac{\sqrt{2+\sqrt{3}}}{2}-\dfrac{2}{\sqrt{6}}+\dfrac{\sqrt{2+\sqrt{3}}}{2\sqrt{3}}\right)\)

\(B=\dfrac{\sqrt{4+2\sqrt{3}}}{2\sqrt{2}}:\left(\dfrac{\sqrt{3}\sqrt{4+2\sqrt{3}}}{2\sqrt{6}}-\dfrac{4}{2\sqrt{6}}+\dfrac{\sqrt{4+2\sqrt{3}}}{2\sqrt{6}}\right)\)

\(B=\dfrac{\sqrt{3}+1}{2\sqrt{2}}:\left(\dfrac{\sqrt{3}\sqrt{4+2\sqrt{3}}-4+\sqrt{4+2\sqrt{3}}}{2\sqrt{6}}\right)\)

\(B=\dfrac{\sqrt{3}+1}{2\sqrt{2}}:\left(\dfrac{\left(\sqrt{3}+1\right)\sqrt{4+2\sqrt{3}}-4}{2\sqrt{6}}\right)\)

\(B=\dfrac{\sqrt{3}+1}{2\sqrt{2}}:\left(\dfrac{\left(\sqrt{3}+1\right)^2-4}{2\sqrt{6}}\right)\)

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

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

\(\Leftrightarrow B=\dfrac{\left(\sqrt{6}+\sqrt{2}\right)}{\left(\sqrt{6}-\sqrt{2}\right)\left(\sqrt{6}+\sqrt{2}\right)}=\dfrac{\sqrt{6}+\sqrt{2}}{4}\)

a: \(=\dfrac{2\sqrt{7}-10-6+\sqrt{7}}{4}+\dfrac{24+6\sqrt{7}-20+5\sqrt{7}}{9}\)

\(=\dfrac{3\sqrt{7}-16}{4}+\dfrac{4+11\sqrt{7}}{9}\)

\(=\dfrac{27\sqrt{7}-144+16+44\sqrt{7}}{36}=\dfrac{71\sqrt{7}-128}{36}\)

b: \(=\dfrac{\sqrt{y}\left(x+y\right)}{\sqrt{xy}}\cdot\dfrac{\sqrt{x}-\sqrt{y}}{x+y}\)

\(=\dfrac{\sqrt{x}-\sqrt{y}}{\sqrt{x}}\)

c: \(=\left(\dfrac{\left(\sqrt{x}-1\right)\left(3\sqrt{x}+1\right)+3\sqrt{x}-1}{\left(3\sqrt{x}-1\right)\left(3\sqrt{x}+1\right)}\right)\cdot\dfrac{3\sqrt{x}-1}{3\sqrt{x}-5}\)

\(=\dfrac{3x+\sqrt{x}-3\sqrt{x}-1+3\sqrt{x}-1}{3\sqrt{x}+1}\cdot\dfrac{1}{3\sqrt{x}-5}\)

\(=\dfrac{3x+\sqrt{x}-2}{\left(3\sqrt{x}+1\right)}\cdot\dfrac{1}{3\sqrt{x}-5}\)

\(=\dfrac{3x+\sqrt{x}-2}{\left(3\sqrt{x}+1\right)\left(3\sqrt{x}-5\right)}\)

a: \(A=6-3\sqrt{3}+4+\sqrt{3}+2\sqrt{3}=10\)

b: \(B=\sqrt{x}-\sqrt{y}-\sqrt{x}-\sqrt{y}=-2\sqrt{y}\)

c: \(C=\dfrac{\sqrt{3}-1}{\sqrt{6}-\sqrt{2}}=\dfrac{1}{\sqrt{2}}=\dfrac{\sqrt{2}}{2}\)

\(\frac{\sqrt{2}-1}{\sqrt{2}+2}-\frac{1}{1+\sqrt{2}}+\frac{\sqrt{2}+1}{\sqrt{2}}=\frac{\sqrt{2}-1}{\sqrt{2}+2}-\frac{\sqrt{2}}{\left(1+\sqrt{2}\right)\sqrt{2}}+\frac{\left(\sqrt{2}+1\right)^2}{\sqrt{2}\left(\sqrt{2}+1\right)}=\frac{\sqrt{2}-1}{2+\sqrt{2}}-\frac{\sqrt{2}}{2+\sqrt{2}}+\frac{3+2\sqrt{2}}{2+\sqrt{2}}=\frac{\sqrt{2}-1-\sqrt{2}+3+2\sqrt{2}}{2+\sqrt{2}}=\frac{2+2\sqrt{2}}{2+\sqrt{2}}\) \(b,\sqrt{x}-2+\frac{10-x}{\sqrt{x}+2}=\left(\sqrt{x}-2\right)+\frac{10-x}{\sqrt{x}+2}=\frac{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)+10-x}{\sqrt{x}+2}=\frac{x-4+10-x}{\sqrt{x}+2}=\frac{6}{\sqrt{x}+2}\)

\(c,\frac{x\sqrt{x}-y\sqrt{y}}{\sqrt{x}-\sqrt{y}}=\frac{\sqrt{x^3}-\sqrt{y^3}}{\sqrt{x}-\sqrt{y}}=\frac{\left(\sqrt{x}\right)^3-\left(\sqrt{y}\right)^3}{\sqrt{x}-\sqrt{y}}=\frac{\left(\sqrt{x}-\sqrt{y}\right)\left(x+\sqrt{xy}+y\right)}{\sqrt{x}-\sqrt{y}}=x+\sqrt{xy}+y\)