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$Q.13 \\ to \: prove \: that : \\ \frac{a {}^{ - 1} }{a {}^{ - 1} + b {}^{ - 1} } + \frac{a {}^{ - 1} }{a {}^{ - 1} - b {}^{ - 1} } = \frac{2b {}^{2} }{b {}^{2} - a {}^{2} } \\ \\ LHS = \frac{a {}^{ - 1} }{a {}^{ - 1} + b {}^{ - 1} } + \frac{a {}^{ - 1} }{a {}^{ - 1} - b {}^{ - 1} } \\ \\ = a {}^{ - 1}( \: \frac{1}{a {}^{ - 1} + b {}^{ - 1} } + \frac{1}{a {}^{ - 1} - b {}^{ - 1} } \: ) \\ \\ a {}^{ - 1} ( \: \frac{a {}^{ - 1} + b {}^{ - 1} + a {}^{ - 1} - b {}^{ - 1} }{(a {}^{ - 1} + b {}^{ - 1})(a {}^{ - 1} - b {}^{ - 1}) } \: ) \\ \\ a {}^{ - 1} ( \: \frac{2a {}^{ - 1} }{a {}^{ - 2} - b {}^{ - 2} \: } \: ) \\ \\ = \frac{2a {}^{ - 2} }{a {}^{ - 2} - b {}^{ - 2} } \\ \\ = \frac{ \frac{2}{a {}^{2} } }{ \frac{1}{a {}^{2} } - \frac{1}{b {}^{2} } } \\ \\ = \frac{ \frac{ \frac{2}{a {}^{2} } }{b {}^{2} - a {}^{2} } }{a {}^{2} b {}^{2} } \\ \\ = \frac{2b {}^{2} }{b {}^{2} - a {}^{2} } \\ \\ = RHS$

$Q.15 \\ to \: find :x {}^{3} - 2x {}^{2} - 7x + 5 \\ \\ given \: that : \\ x = \frac{1}{2 - \sqrt{ 3 } } \\ \\ = \frac{1}{2 - \sqrt{3} } \times \frac{2 + \sqrt{3} }{2 + \sqrt{3} } \\ \\ = \frac{2 + \sqrt{3} }{2 {}^{2} - 3 } \\ \\ x= 2 + \sqrt{3} \\ \\ substituting \: value \: of \: x \: in \\ x {}^{3} - 2x {}^{2} - 7x + 5 \\ we \: get \\ \\ x {}^{3} - 2x {}^{2} - 7x + 5 = x {}^{2} (x - 2) - 7x + 5 \\ \\ = (2 + \sqrt{3} ) {}^{2} (2 + \sqrt{3} - 2) - 7(2 + \sqrt{3} ) + 5 \\ \\ = (7 + 4 \sqrt{3} )( \sqrt{3} \: ) - 14 - 7 \sqrt{3} + 5 \\ \\ = 7 \sqrt{3} + 12 - 14 - 7 \sqrt{3} + 5 \\ \\ = 17 - 14 \\ \\ = 3$