Question

Kindly answer the following questions :-

Answer 1

$Given \: \: Questions \: \: Are \: \: \\ \\ QUESTION \: \: no \: 18 \\ \\ Answer \: \\ \\ 18) \: \: \: x = \frac{ \sqrt{3} + \sqrt{2} }{ \sqrt{3} - \sqrt{2} } \\ \\ Multiply \: \: Numerator \: And \: Denominator \: by \: \\ \sqrt{3} + \sqrt{2} \: \: we \: have \\ \\ x = \frac{ \sqrt{3} + \sqrt{2} }{ \sqrt{3} - \sqrt{2} } \times \frac{ \sqrt{3} + \sqrt{2} }{ \sqrt{3} + \sqrt{2} } \\ \\ x = \frac{( \sqrt{3} + \sqrt{2} ) {}^{2} }{( \sqrt{3}) {}^{2} - ( \sqrt{2} ) {}^{2} } \\ \\ x = \frac{3 + 2 + 2 \sqrt{6} }{ 3 - 2} \\ \\ x = 5 + 2 \sqrt{6} \\ \\ x {}^{2} = (5 + 2 \sqrt{6} ) {}^{2} \\ \\ x {}^{2} = 25 + 24 + 20 \sqrt{6} \\ \\ x {}^{2} = 49 + 20 \sqrt{6} \: \: \: ... \: i\\ \\ \frac{1}{x {}^{2} } = \frac{1}{49 + 20 \sqrt{6} } \\ \\ Multiply \: Numerator \: \: and \: \\ Denominator \: by \: \: 49 - 20 \sqrt{6} \\ \\ \frac{1}{x {}^{2} } = \frac{1}{49 + 20 \sqrt{6} } \times \frac{49 - 20 \sqrt{6} }{49 - 20 \sqrt{6} } \\ \\ \frac{1}{x {}^{2} } = \frac{49 - 20 \sqrt{6} }{(49) {}^{2} - (20 \sqrt{6}) {}^{2} } \\ \\ \frac{1}{x {}^{2} } = \frac{49 - 20 \sqrt{6} }{2401 - 2400} \\ \\ \frac{1}{x {}^{2} } = 49 - 20 \sqrt{6} \: \: ...ii \\ \\ from \: i \: and \: ii \: we \: have \\ \\ x {}^{2} + \frac{1}{x {}^{2} } = 49 + 20 \sqrt{6} + 49 - 20 \sqrt{6} \\ \\ x {}^{2} + \frac{1}{x {}^{2} } = 98 \\ \\ Question \: Number \: \: 20 \\ \\ Answer \: \: \\ \\ a {}^{x} = b {}^{y} = c {}^{z} = k \: \: \: \: let \\ \\ a = k {}^{ \frac{1}{x} } \: \: \: .... \: \: i \\ \\ b = k {}^{ \frac{1}{y} } \: \: \: ... \: ii \\ \\ c = k {}^{ \frac{1}{z} } \: \: \: .... \: \: \: iii \\ \\ Multiply \: these \: three \: equations \: we \: have \\ \\ abc = k {}^{ \frac{1}{x} } \times k {}^{ \frac{1}{y} } \times k {}^{ \frac{1}{z} } \\ \\ k {}^{( \frac{1}{x} + \frac{1}{y} + \frac{1}{z} )} = 1 \\ \\ k {}^{( \frac{1}{x} + \frac{1}{y} + \frac{1}{z} )} = k {}^{0} \\ \\ now \: compare \: powers \: of \: k \: we \: have \\ \\ \frac{1}{x} + \frac{1}{y} + \frac{1}{z} = 0 \\ \\ \frac{zx + zy + xy}{xyz} = 0 \\ \\ xy + yz + zx = 0$