Question

\(\frac{2}{\sqrt{10+2\sqrt{21}}}-\frac{1}{\sqrt{12-2\sqrt{35}}}+\frac{1}{\sqrt{8-2\sqrt{15}}}=\)

• A. \(\sqrt7+\sqrt5\)
• B. \(\sqrt5-\sqrt3\)
• C. 0
• D. 1

Hint:

Expressions of the form \(a\pm2\sqrt b\) can often be written as \((\sqrt m\pm\sqrt n)^2\).

Answer 1

The Correct Option is D

Step 1: Simplifying each radical.
\[ 10+2\sqrt{21} =(\sqrt7+\sqrt3)^2 \] \[ \sqrt{10+2\sqrt{21}} =\sqrt7+\sqrt3 \] Similarly, \[ \sqrt{12-2\sqrt{35}} =\sqrt7-\sqrt5 \] \[ \sqrt{8-2\sqrt{15}} =\sqrt5-\sqrt3 \]

Step 2: Substituting into the expression.
\[ \frac{2}{\sqrt7+\sqrt3} -\frac{1}{\sqrt7-\sqrt5} +\frac{1}{\sqrt5-\sqrt3} \] Rationalizing, \[ \frac{2}{\sqrt7+\sqrt3} =\frac{\sqrt7-\sqrt3}{2} \] \[ \frac{1}{\sqrt7-\sqrt5} =\frac{\sqrt7+\sqrt5}{2} \] \[ \frac{1}{\sqrt5-\sqrt3} =\frac{\sqrt5+\sqrt3}{2} \]

Step 3: Adding the terms.
\[ \frac{\sqrt7-\sqrt3-\sqrt7-\sqrt5+\sqrt5+\sqrt3}{2} \] \[ =\frac{2}{2} =1 \] {1}