$$\frac{{3\sqrt 2 }}{{\sqrt 6 + \sqrt 3 }} - $$ $$\frac{{2\sqrt 6 }}{{\sqrt 3 + 1}} + $$ $$\frac{{2\sqrt 3 }}{{\sqrt 6 + 2}}$$ is equal to = ?
A. 3
B. 2
C. 0
D. $$\sqrt 3 $$
Answer: Option C
Solution(By Examveda Team)
$$\frac{{3\sqrt 2 }}{{\sqrt 6 + \sqrt 3 }} - \frac{{2\sqrt 6 }}{{\sqrt 3 + 1}} + \frac{{2\sqrt 3 }}{{\sqrt 6 + 2}}$$$$ = \left( {\frac{{3\sqrt 2 }}{{\sqrt 6 + \sqrt 3 }} \times \frac{{\sqrt 6 - \sqrt 3 }}{{\sqrt 6 - \sqrt 3 }}{\text{ }}} \right) - $$ $$\left( {\frac{{2\sqrt 6 }}{{\sqrt 3 + 1}} \times \frac{{\sqrt 3 - 1}}{{\sqrt 3 - 1}}{\text{ }}} \right) + $$ $$\left( {\frac{{2\sqrt 3 }}{{\sqrt 6 + 2}}{\text{ }} \times \frac{{\sqrt 6 - 2}}{{\sqrt 6 - 2}}{\text{ }}} \right)$$
$$ = \frac{{3\sqrt 2 \left( {\sqrt 6 - \sqrt 3 } \right)}}{3} - $$ $$\frac{{2\sqrt 6 \left( {\sqrt 3 - 1} \right)}}{2} + $$ $$\frac{{2\sqrt 3 \left( {\sqrt 6 - 2} \right)}}{2}$$
$$\eqalign{ & = \sqrt {12} - \sqrt 6 - \sqrt {18} + \sqrt 6 + \sqrt {18} - 2\sqrt 3 \cr & = \sqrt {12} - 2\sqrt 3 \cr & = 2\sqrt 3 - 2\sqrt 3 \cr & = 0 \cr} $$
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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