${{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 } \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
$0$
Similar Questions
For $x \ne 0,{\left( {{{{x^l}} \over {{x^m}}}} \right)^{({l^2} + lm + {m^2})}}$${\left( {{{{x^m}} \over {{x^n}}}} \right)^{({m^2} + nm + {n^2})}}{\left( {{{{x^n}} \over {{x^l}}}} \right)^{({n^2} + nl + {l^2})}}=$
For $x \ne 0,{\left( {{{{x^l}} \over {{x^m}}}} \right)^{({l^2} + lm + {m^2})}}$${\left( {{{{x^m}} \over {{x^n}}}} \right)^{({m^2} + nm + {n^2})}}{\left( {{{{x^n}} \over {{x^l}}}} \right)^{({n^2} + nl + {l^2})}}=$
The rationalising factor of ${a^{1/3}} + {a^{ - 1/3}}$ is
The rationalising factor of ${a^{1/3}} + {a^{ - 1/3}}$ is
If $x = {2^{1/3}} - {2^{ - 1/3}},$ then $2{x^3} + 6x = $
If $x = {2^{1/3}} - {2^{ - 1/3}},$ then $2{x^3} + 6x = $
If ${a^x} = bc,{b^y} = ca,\,{c^z} = ab,$ then $xyz$=
If ${a^x} = bc,{b^y} = ca,\,{c^z} = ab,$ then $xyz$=
Let ${7 \over {{2^{1/2}} + {2^{1/4}} + 1}}$$ = A + B{.2^{1/4}} + C{.2^{1/2}} + D{.2^{3/4}}$, then $A+B+C+D= . . .$
Let ${7 \over {{2^{1/2}} + {2^{1/4}} + 1}}$$ = A + B{.2^{1/4}} + C{.2^{1/2}} + D{.2^{3/4}}$, then $A+B+C+D= . . .$