a) +) \(\displaystyle {5 \over {3\sqrt 8 }} = {{5\sqrt 8 } \over {3\sqrt 8 .\sqrt 8 }} = {{5\sqrt 8 } \over {3.8}} = {5 \over {24}}\sqrt 8 \)
+) \(\displaystyle {2 \over {\sqrt b }} = {{2\sqrt b } \over {\sqrt b .\sqrt b }} = {2 \over b}\sqrt b \)
b) \(\displaystyle {5 \over {5 - 2\sqrt 3 }} = {{5\left( {5 + 2\sqrt 3 } \right)} \over {\left( {5 - 2\sqrt 3 } \right)\left( {5 + 2\sqrt 3 } \right)}} \\ \displaystyle = {{5\left( {5 + 2\sqrt 3 } \right)} \over {25 - 12}} = {{5\left( {5 + 2\sqrt 3 } \right)} \over {13}}\)
\(\displaystyle {{2a} \over {1 - \sqrt a }} = {{2a\left( {1 + \sqrt a } \right)} \over {\left( {1 - \sqrt a } \right)\left( {1 + \sqrt a } \right)}}\\ \displaystyle = {{2a\left( {1 + \sqrt a } \right)} \over {1 - a}}\)
c) \(\displaystyle {4 \over {\sqrt 7 + \sqrt 5 }} = {{4\left( {\sqrt 7 - \sqrt 5 } \right)} \over {\left( {\sqrt 7 + \sqrt 5 } \right)\left( {\sqrt 7 - \sqrt 5 } \right)}} \\ \displaystyle = {{4\left( {\sqrt 7 - \sqrt 5 } \right)} \over {7 - 5}} = 2\left( {\sqrt 7 - \sqrt 5 } \right)\)
\(\displaystyle {{6a} \over {2\sqrt a - \sqrt b }} = {{6a\left( {2\sqrt a + \sqrt b } \right)} \over {\left( {2\sqrt a - \sqrt b } \right)\left( {2\sqrt a + \sqrt b } \right)}} \\ \displaystyle = {{6a\left( {2\sqrt a + \sqrt b } \right)} \over {4a - b}}\)
