Bài 19:
a: \(\dfrac{1}{\sqrt{3}+\sqrt{2}-1}=\dfrac{\sqrt{3}+\sqrt{2}+1}{\left(\sqrt{3}+\sqrt{2}-1\right)\left(\sqrt{3}+\sqrt{2}+1\right)}\)
\(=\dfrac{\sqrt{3}+\sqrt{2}+1}{5+2\sqrt{6}-1}=\dfrac{\sqrt{3}+\sqrt{2}+1}{4+2\sqrt{6}}\)
\(=\dfrac{\left(\sqrt{3}+\sqrt{2}+1\right)\left(2\sqrt{6}-4\right)}{\left(2\sqrt{6}+4\right)\left(2\sqrt{6}-4\right)}\)
\(=\dfrac{2\sqrt{18}-4\sqrt{3}+2\sqrt{12}-4\sqrt{2}+2\sqrt{6}-4}{24-16}\)
\(=\dfrac{6\sqrt{2}-4\sqrt{3}+4\sqrt{3}-4\sqrt{2}+2\sqrt{6}-4}{8}\)
\(=\dfrac{2\sqrt{2}+2\sqrt{6}-4}{8}=\dfrac{\sqrt{2}+\sqrt{6}-2}{4}\)
b:
\(\dfrac{1}{\sqrt{14}-\sqrt{6+\sqrt{35}}}\)
\(=\dfrac{\sqrt{2}}{\sqrt{28}-\sqrt{12+2\sqrt{35}}}\)
\(=\dfrac{\sqrt{2}}{2\sqrt{7}-\left(\sqrt{7}+\sqrt{5}\right)}=\dfrac{\sqrt{2}}{\sqrt{7}-\sqrt{5}}\)
\(=\dfrac{\sqrt{2}\left(\sqrt{7}+\sqrt{5}\right)}{7-5}=\dfrac{\sqrt{14}+\sqrt{10}}{2}\)
Bài 20:
a: \(A=\dfrac{\sqrt{6}+\sqrt{14}}{2\sqrt{3}-\sqrt{7}}=\dfrac{\left(\sqrt{6}+\sqrt{14}\right)\left(2\sqrt{3}+\sqrt{7}\right)}{\left(2\sqrt{3}-\sqrt{7}\right)\left(2\sqrt{3}+\sqrt{7}\right)}\)
\(=\dfrac{2\sqrt{18}+\sqrt{42}+2\sqrt{42}+\sqrt{98}}{12-7}\)
\(=\dfrac{6\sqrt{2}+7\sqrt{2}+3\sqrt{42}}{5}=\dfrac{13\sqrt{2}+3\sqrt{42}}{5}\)
b: \(B=\dfrac{3+4\sqrt{3}}{\sqrt{6}+\sqrt{2}-\sqrt{5}}\)
\(=\dfrac{\left(3+4\sqrt{3}\right)\left(\sqrt{6}+\sqrt{2}+\sqrt{5}\right)}{\left(\sqrt{6}+\sqrt{2}-\sqrt{5}\right)\left(\sqrt{6}+\sqrt{2}+\sqrt{5}\right)}\)
\(=\dfrac{\left(3+4\sqrt{3}\right)\left(\sqrt{6}+\sqrt{2}+\sqrt{5}\right)}{8+2\sqrt{12}-5}\)
\(=\dfrac{\left(3+4\sqrt{3}\right)\left(\sqrt{6}+\sqrt{2}+\sqrt{5}\right)}{3+4\sqrt{3}}=\sqrt{6}+\sqrt{2}+\sqrt{5}\)
c: \(C=\dfrac{1}{2+\sqrt{5}+2\sqrt{2}+\sqrt{10}}\)
\(=\dfrac{1}{2+\sqrt{5}+\sqrt{2}\left(2+\sqrt{5}\right)}\)
\(=\dfrac{1}{\left(\sqrt{2}+1\right)\left(\sqrt{5}+2\right)}=\dfrac{1}{\sqrt{2}+1}\cdot\dfrac{1}{\sqrt{5}+2}\)
\(=\left(\sqrt{2}-1\right)\left(\sqrt{5}-2\right)=\sqrt{10}-2\sqrt{2}-\sqrt{5}+2\)
