Answer:
[tex]x = \sqrt{7} + \sqrt{3} \\ x - \frac{4}{x} = 2 \sqrt{3} \\ ( \sqrt{7} + \sqrt{3} ) - \frac{4}{( \sqrt{7} + \sqrt{3}) } = 2 \sqrt{3} \\ \frac{ {( \sqrt{7} + \sqrt{3}) }^{2} - 4}{ \sqrt{7} + \sqrt{3} } = 2 \sqrt{3} \\ \frac{ {( \sqrt{7} )}^{2} + {( \sqrt{3} )}^{2} + 2( \sqrt{7})( \sqrt{3} ) - 4}{ \sqrt{7} + \sqrt{3} } = 2 \sqrt{3} \\ \frac{7 + 3 + 2 \sqrt{21} - 4}{ \sqrt{7} + \sqrt{3} } = 2 \sqrt{3} \\ \frac{6 + 2 \sqrt{21} }{ \sqrt{7} + \sqrt{3} } = 2 \sqrt{3} \\ \frac{2 \times \sqrt{3} \times \sqrt{3} + 2 \times \sqrt{7} \times \sqrt{3} }{ \sqrt{7} + \sqrt{3} } = 2 \sqrt{3} \\ \frac{2 \sqrt{3} ( \sqrt{7} + \sqrt{3} )}{ \sqrt{7} + \sqrt{3} } = 2 \sqrt{3} \\ 2 \sqrt{3} = 2 \sqrt{3} \\ LHS=RHS \\ Hence \: proved[/tex]
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Answer:
[tex]x = \sqrt{7} + \sqrt{3} \\ x - \frac{4}{x} = 2 \sqrt{3} \\ ( \sqrt{7} + \sqrt{3} ) - \frac{4}{( \sqrt{7} + \sqrt{3}) } = 2 \sqrt{3} \\ \frac{ {( \sqrt{7} + \sqrt{3}) }^{2} - 4}{ \sqrt{7} + \sqrt{3} } = 2 \sqrt{3} \\ \frac{ {( \sqrt{7} )}^{2} + {( \sqrt{3} )}^{2} + 2( \sqrt{7})( \sqrt{3} ) - 4}{ \sqrt{7} + \sqrt{3} } = 2 \sqrt{3} \\ \frac{7 + 3 + 2 \sqrt{21} - 4}{ \sqrt{7} + \sqrt{3} } = 2 \sqrt{3} \\ \frac{6 + 2 \sqrt{21} }{ \sqrt{7} + \sqrt{3} } = 2 \sqrt{3} \\ \frac{2 \times \sqrt{3} \times \sqrt{3} + 2 \times \sqrt{7} \times \sqrt{3} }{ \sqrt{7} + \sqrt{3} } = 2 \sqrt{3} \\ \frac{2 \sqrt{3} ( \sqrt{7} + \sqrt{3} )}{ \sqrt{7} + \sqrt{3} } = 2 \sqrt{3} \\ 2 \sqrt{3} = 2 \sqrt{3} \\ LHS=RHS \\ Hence \: proved[/tex]
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