Question

simplify 1/√3+√2-2/√5-√3-√3/√2-√5

Answer 1

Hey friend,
Here is the answer you were looking for:
$\frac{1}{ \sqrt{3} + \sqrt{2} } - \frac{2}{ \sqrt{5} - \sqrt{3} } - \frac{ \sqrt{3} }{ \sqrt{2} - \sqrt{5} } \\ \\ on \: rationalizing \: the \: denominator \: we \: get \\ \\ = \frac{1}{ \sqrt{3} + \sqrt{2} } \times \frac{ \sqrt{3} - \sqrt{2} }{ \sqrt{3} - \sqrt{2} } - \frac{2}{ \sqrt{5} - \sqrt{3} } \times \frac{ \sqrt{5} + \sqrt{3} }{ \sqrt{5} + \sqrt{3} } - \frac{ \sqrt{3} }{ \sqrt{2} - \sqrt{5} } \times \frac{ \sqrt{2} + \sqrt{5} }{ \sqrt{2} + \sqrt{5} } \\ \\ using \: the \: identity \\ (a + b)(a - b) = {a}^{2} - {b}^{2} \\ \\ = \frac{ \sqrt{3} - \sqrt{2} }{ {( \sqrt{3} )}^{2} - {( \sqrt{2} )}^{2} } - \frac{2 \times \sqrt{5} + 2 \times \sqrt{3} }{ {( \sqrt{5} )}^{2} - {( \sqrt{3} )}^{2} } - \frac{ \sqrt{3} \times \sqrt{2} + \sqrt{3} \times \sqrt{5} }{ {( \sqrt{2} )}^{2} - {( \sqrt{5} )}^{2} } \\ \\ = \frac{ \sqrt{3} - \sqrt{2} }{3 - 2} - \frac{2 \sqrt{5} + 2 \sqrt{3} }{5 - 3} - \frac{ \sqrt{6} + \sqrt{15} }{2 - 5} \\ \\ = \sqrt{3} - \sqrt{2} - \frac{2 \sqrt{5} + 2 \sqrt{3} }{2} - \frac{ \sqrt{6} + \sqrt{15} }{ - 3} \\ \\ = \sqrt{3} - \sqrt{2} - \frac{2 \sqrt{5} + 2 \sqrt{3} }{2} - \frac{ - \sqrt{6} - \sqrt{15} }{3} \\ \\ = \frac{ \sqrt{3} \times 6 - \sqrt{2} \times 6 - 2 \sqrt{5} \times 3 - 2 \sqrt{3} \times 3 + \sqrt{6} \times 2 + \sqrt{15} \times 2}{6} \\ \\ = \frac{6 \sqrt{3} - 3 \sqrt{2} - 6 \sqrt{5} - 6 \sqrt{3} + 2 \sqrt{6} + 2 \sqrt{15} }{6} \\ \\ = \frac{ - 3 \sqrt{2} - 6 \sqrt{5} + 2 \sqrt{6} + 2 \sqrt{15} }{6}$

Hope this helps!!!!

@Mahak24

Thanks...
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