Question

Prove that : (3√2/√3-√2) - (4√3/√6+√2) + (√6/√2+√3) =0​

Answer 1

Step-by-step explanation:

$\frac{3 \sqrt{2} }{ \sqrt{3} - \sqrt{2} } - \frac{4 \sqrt{3} }{ \sqrt{6} + \sqrt{2} } + \frac{ \sqrt{6} }{ \sqrt{2} + \sqrt{3} } = 0 \\ Taking \: L.H.S: \\ = (\frac{3 \sqrt{2} }{ \sqrt{3} - \sqrt{2} } \times \frac{ \sqrt{3} + \sqrt{2} }{ \sqrt{3} + \sqrt{2} } ) - (\frac{4 \sqrt{3} }{ \sqrt{6} + \sqrt{2} } \times \frac{ \sqrt{6} - \sqrt{2} }{ \sqrt{6} - \sqrt{2} }) + ( \frac{ \sqrt{6} }{ \sqrt{2} + \sqrt{3} } \times \frac{ \sqrt{2} - \sqrt{3} }{ \sqrt{2} - \sqrt{3} } ) \\ = \frac{ 3\sqrt{2}( \sqrt{3} + \sqrt{2} ) }{ {( \sqrt{3} )}^{2} - {( \sqrt{2} )}^{2} } - \frac{4 \sqrt{3} ( \sqrt{6} - \sqrt{2} )}{ {( \sqrt{6} )}^{2} - {( \sqrt{2} )}^{2} } + \frac{ \sqrt{6} ( \sqrt{2} - \sqrt{3} )}{ {( \sqrt{2} )}^{2} - {( \sqrt{3} )}^{2} } \\ = \frac{3 \sqrt{6} + 3 \sqrt{4} }{3 - 2} - \frac{4 \sqrt{18} + 4 \sqrt{6} }{6 - 2} + \frac{ \sqrt{12} - \sqrt{18} }{2 - 3} \\ = 3 \sqrt{6} + 3\sqrt{4} - \frac{4 \sqrt{18} + 4 \sqrt{6} }{4} - \sqrt{12} + \sqrt{18} \\ = \frac{12 \sqrt{6} + 12 \sqrt{4} - 4 \sqrt{18} + 4 \sqrt{6} - 4 \sqrt{12} + 4 \sqrt{18} }{4} \\ = \frac{12 \sqrt{6} + 12\sqrt{2 \times 2} - 4 \sqrt{3 \times 3 \times 2} + 4 \sqrt{6} - 4 \sqrt{2 \times 2 \times 3} + 4 \sqrt{3 \times 3 \times 2} }{4} \\ = \frac{12 \sqrt{6} +12 \times 2 - 4 \times 3 \sqrt{2} + 4 \sqrt{6} - 4 \times 2 \sqrt{3} + 4 \times 3 \sqrt{2} }{4} \\ = \frac{12 \sqrt{6} + 24 - 12 \sqrt{2} + 4 \sqrt{6} - 8 \sqrt{3} + 12 \sqrt{2} }{4} \\ = \frac{16 \sqrt{6} - 8 \sqrt{3} + 24}{4} \\ = \frac{4(4 \sqrt{6} - 2 \sqrt{3} + 6 )}{4} \\ = 4 \sqrt{6} - 2 \sqrt{3} + 6 \: .is \: not \: equal \: to \: 0$

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