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Basic of Logarithms
easy
${{3\sqrt 2 } \over {\sqrt 6 + \sqrt 3 }} - {{4\sqrt 3 } \over {\sqrt 6 + \sqrt 2 }} + {{\sqrt 6 } \over {\sqrt 3 + \sqrt 2 }} = $
A
$5\sqrt 2 $
B
$3\sqrt 2 $
C
$2\sqrt 3 $
D
$0$
Solution
(d) ${{3\sqrt 2 } \over {\sqrt 6 + \sqrt 3 }} – {{4\sqrt 3 } \over {\sqrt 6 + \sqrt 2 }} + {{\sqrt 6 } \over {\sqrt 3 + \sqrt 2 }}$
$ = {{3\sqrt 2 (\sqrt 6 – \sqrt 3 )} \over {6 – 3}} – {{4\sqrt 3 \,(\sqrt 6 – \sqrt 2 )} \over {6 – 2}} + {{\sqrt 6 (\sqrt 3 – \sqrt 2 )} \over {3 – 2}}$
$ = \sqrt 2 \,(\sqrt 6 – \sqrt 3 ) – \sqrt 3 \,(\sqrt 6 – \sqrt 2 ) + \sqrt 6 (\sqrt 3 – \sqrt 2 )= 0.$
Standard 11
Mathematics
