InterviewSolution
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1). \(\sqrt {6\;}+ \sqrt {12} \)2). \(\sqrt 3+ \sqrt {12} \)3). \(\sqrt 2+ \sqrt {12} \)4). 0 |
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Answer» CONJUGATE of (√a + √b) is (√a - √b) and vice versa. $(\begin{array}{l}\left[ {\frac{{3\sqrt 2 }}{{\sqrt 5+ \sqrt 8 }} - \frac{{4\sqrt 3 }}{{2 + \sqrt 8 }} + \frac{{\sqrt 8 }}{{\sqrt 2+ \sqrt 3 }} + \frac{4}{{\sqrt {10}- \sqrt 6 }}} \right] = \left[ {\frac{{3\sqrt 2 \left( {\sqrt 8- \sqrt 5 } \right)}}{{8 - 5}} - \frac{{4\sqrt 3 \left( {\sqrt 8- 2} \right)}}{{8 - 4}} + \frac{{\sqrt 8 \left( {\sqrt 3- \sqrt 2 } \right)}}{{3 - 2}} + \frac{{4\left( {\sqrt {10}+ \sqrt 6 } \right)}}{{10 - 6}}} \right]\\ = \sqrt {16}- \sqrt {10}- \sqrt {24}+ \sqrt {12}+ \sqrt {24}- \sqrt {16}+ \sqrt {10}+ \sqrt 6= \sqrt {6\;}+ \sqrt {12}\END{array})$ |
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