Question

Simplify root3-1/2root2+root3

Answer 1

Explanation:

$\frac{\sqrt3-1}{2\sqrt2+\sqrt3}$

$= \frac{\sqrt3 - 1}{2\sqrt2+\sqrt3} \times \frac{(2\sqrt2 - \sqrt3)}{(2\sqrt2 - \sqrt3)}$

$=\frac{(\sqrt3 -1)(2\sqrt2-\sqrt3)}{(2\sqrt2)^2-(\sqrt3)^2}$ [∵$(a+b)(a-b)=a^2 - b^2$]

$=\frac{(\sqrt3-1)(2\sqrt2-\sqrt3)}{8-3} = \frac{(\sqrt3-1)(2\sqrt2-\sqrt3)}{5}$

$=\frac{\sqrt3(2\sqrt2-\sqrt3) - 1(2\sqrt2-\sqrt3)}{5}$

$=\frac {2\sqrt6 - 3 - 2\sqrt2 + \sqrt3}{5}$

Answer: 2√6 - 3 - 2√2 + √3/5

Answer 2

Answer:

Hope it helps you wishing you a good luck for your future