$$\left[ {\frac{{\sqrt 3 + \sqrt 2 }}{{\sqrt 3 - \sqrt 2 }} - \frac{{\sqrt 3 - \sqrt 2 }}{{\sqrt 3 + \sqrt 2 }}} \right]$$ simplifies to = ?
A. $${\text{2}}\sqrt 6 $$
B. $${\text{4}}\sqrt 6 $$
C. $${\text{2}}\sqrt 3 $$
D. $${\text{3}}\sqrt 2 $$
Answer: Option B
Solution(By Examveda Team)
$$\left[ {\frac{{\sqrt 3 + \sqrt 2 }}{{\sqrt 3 - \sqrt 2 }} - \frac{{\sqrt 3 - \sqrt 2 }}{{\sqrt 3 + \sqrt 2 }}} \right]$$$$ = \frac{{\sqrt 3 + \sqrt 2 }}{{\sqrt 3 - \sqrt 2 }} \times $$ $$\frac{{\sqrt 3 + \sqrt 2 }}{{\sqrt 3 + \sqrt 2 }} - $$ $$\frac{{\sqrt 3 - \sqrt 2 }}{{\sqrt 3 + \sqrt 2 }} \times $$ $$\frac{{\sqrt 3 - \sqrt 2 }}{{\sqrt 3 - \sqrt 2 }}$$
$$\eqalign{ & = \frac{{{{\left( {\sqrt 3 + \sqrt 2 } \right)}^2}}}{{3 - 2}} - \frac{{{{\left( {\sqrt 3 - \sqrt 2 } \right)}^2}}}{{3 - 2}} \cr & = \left( {3 + 2 + 2\sqrt 6 } \right) - \left( {3 + 2 - 2\sqrt 6 } \right) \cr & = 4\sqrt 6 {\text{ }} \cr} $$
Related Questions on Surds and Indices
A. $$\frac{1}{2}$$
B. 1
C. 2
D. $$\frac{7}{2}$$
Given that 100.48 = x, 100.70 = y and xz = y2, then the value of z is close to:
A. 1.45
B. 1.88
C. 2.9
D. 3.7
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