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a=1/(√3+√2+1)=(√3-(√2+1)/[3-(√2+1)^2]
=(√3-√2-1)/(3-(3+2√2)
=(√3-√2-1)/(-2√2)
=-(√6-2-√2)/4
=(2+√2-√6)/4
\(\dfrac{2ab}{\sqrt{a}-\sqrt{b}}=\dfrac{2ab\left(\sqrt{a}+\sqrt{b}\right)}{\left(\sqrt{a}-\sqrt{b}\right)\left(\sqrt{a}+\sqrt{b}\right)}=\dfrac{2ab\left(\sqrt{a}+\sqrt{b}\right)}{a-b}\)
\(\dfrac{1}{\sqrt{x}-\sqrt{y}}=\dfrac{\sqrt{x}+\sqrt{y}}{\left(\sqrt{x}-\sqrt{y}\right)\left(\sqrt{x}+\sqrt{y}\right)}=\dfrac{\sqrt{x}+\sqrt{y}}{x-y}\)
\(\dfrac{3}{\sqrt{10}+\sqrt{7}}=\dfrac{3\left(\sqrt{10}-\sqrt{7}\right)}{\left(\sqrt{10}+\sqrt{7}\right)\left(\sqrt{10}-\sqrt{7}\right)}=\dfrac{3\left(\sqrt{10}-\sqrt{7}\right)}{10-7}=\dfrac{3\left(\sqrt{10}-\sqrt{7}\right)}{3}=\sqrt{10}-\sqrt{7}\)
\(\dfrac{2}{\sqrt{6}-\sqrt{5}}=\dfrac{2\left(\sqrt{6}+\sqrt{5}\right)}{\left(\sqrt{6}-\sqrt{5}\right)\left(\sqrt{6}+\sqrt{5}\right)}=\dfrac{2\left(\sqrt{6}+\sqrt{5}\right)}{6-5}=2\left(\sqrt{6}+\sqrt{5}\right)\)
a: \(=\dfrac{\sqrt{5}+\sqrt{2}+\sqrt{3}}{7+2\sqrt{10}-3}=\dfrac{\sqrt{5}+\sqrt{2}+\sqrt{3}}{4+2\sqrt{10}}\)
\(=\dfrac{-\left(\sqrt{3}+\sqrt{2}+\sqrt{5}\right)\left(4-2\sqrt{10}\right)}{24}\)
b: \(=\dfrac{2+\sqrt{3}+\sqrt{5}}{4-8+2\sqrt{15}}=\dfrac{2+\sqrt{3}+\sqrt{5}}{2\sqrt{15}-4}\)
\(=\dfrac{\left(2+\sqrt{3}+\sqrt{5}\right)\left(2\sqrt{15}+4\right)}{44}\)
a. \(\dfrac{1}{\sqrt{5}-\sqrt{3}+\sqrt{2}}=\dfrac{\sqrt{5}-\sqrt{3}-\sqrt{2}}{\left(\sqrt{5}-\sqrt{3}+\sqrt{2}\right)\left(\sqrt{5}-\sqrt{3}-\sqrt{2}\right)}=\dfrac{\sqrt{5}-\sqrt{3}-\sqrt{2}}{\left(\sqrt{5}-\sqrt{3}\right)^2-2}=\dfrac{\sqrt{5}-\sqrt{3}-\sqrt{2}}{5+3-2-2\sqrt{15}}=\dfrac{\sqrt{5}-\sqrt{3}-\sqrt{2}}{6-2\sqrt{15}}=\dfrac{\left(\sqrt{5}-\sqrt{3}-\sqrt{2}\right)\left(3+\sqrt{15}\right)}{\left(3-\sqrt{15}\right)\left(3+\sqrt{15}\right)2}=\dfrac{3\sqrt{5}-3\sqrt{3}-3\sqrt{2}+5\sqrt{3}-3\sqrt{5}-\sqrt{30}}{\left(9-15\right).2}=\dfrac{2\sqrt{3}-3\sqrt{2}-\sqrt{30}}{-12}\)b. \(\dfrac{1}{2-\sqrt{3}-\sqrt{5}}=\dfrac{2-\sqrt{3}+\sqrt{5}}{\left(2-\sqrt{3}-\sqrt{5}\right)\left(2-\sqrt{3}+\sqrt{5}\right)}=\dfrac{2-\sqrt{3}+\sqrt{5}}{\left(2-\sqrt{3}\right)^2-5}=\dfrac{2-\sqrt{3}+\sqrt{5}}{4-4\sqrt{3}+3-5}=\dfrac{2-\sqrt{3}+\sqrt{5}}{2-4\sqrt{3}}=\dfrac{\left(2-\sqrt{3}+\sqrt{5}\right)\left(1+2\sqrt{3}\right)}{2\left(1-2\sqrt{3}\right)\left(1+2\sqrt{3}\right)}=\dfrac{2+4\sqrt{3}-\sqrt{3}-6+\sqrt{5}+2\sqrt{15}}{2.\left(1-12\right)}=\dfrac{3\sqrt{3}+\sqrt{5}+2\sqrt{15}-4}{-22}\)
a) \(\dfrac{1}{\sqrt{3}+\sqrt{2}+1}\) = \(\dfrac{\sqrt{3}+1-\sqrt{2}}{\left(\sqrt{3}+1+\sqrt{2}\right)\left(\sqrt{3}+1-\sqrt{2}\right)}\)
= \(\dfrac{\sqrt{3}+1-\sqrt{2}}{\left(\sqrt{3}+1\right)^2-2}=\dfrac{\left(\sqrt{3}+1-\sqrt{2}\right)\left(\sqrt{3}-1\right)}{2\left(\sqrt{3}+1\right)\left(\sqrt{3}-1\right)}\)
= \(\dfrac{3-\sqrt{3}+\sqrt{3}-1-\sqrt{6}+\sqrt{2}}{2\left(3-1\right)}\) = \(\dfrac{2-\sqrt{6}+\sqrt{2}}{4}\)
b) \(\dfrac{1}{\sqrt{5}+2-\sqrt{3}}=\dfrac{\sqrt{5}+2+\sqrt{3}}{\left(\sqrt{5}+2\right)^2-3}\) = \(\dfrac{\sqrt{5}+\sqrt{3}+2}{4\sqrt{5}+6}\)
= \(\dfrac{\left(\sqrt{5}+\sqrt{3}+2\right)\left(4\sqrt{5}-6\right)}{\left(4\sqrt{5}+6\right)\left(4\sqrt{5}-6\right)}\) = \(\dfrac{20-6\sqrt{5}+4\sqrt{15}-6\sqrt{3}+8\sqrt{5}-12}{\left(4\sqrt{5}\right)^2-36}\)
= \(\dfrac{8+2\sqrt{5}-6\sqrt{3}+4\sqrt{15}}{44}\) = \(\dfrac{2\left(4+\sqrt{5}-3\sqrt{3}+2\sqrt{15}\right)}{2\left(22\right)}\)
= \(\dfrac{4+\sqrt{5}-3\sqrt{3}+2\sqrt{15}}{22}\)
\(\dfrac{1}{1-\sqrt{2}}=-1-\sqrt{2}\)