Question

solve these problems please​

Answer 1

Solution:-

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$1. \frac{3 + \sqrt{2} }{3 - \sqrt{2} } = a + b \sqrt{2} \\ \\ = \frac{3 + \sqrt{2} }{3 - \sqrt{2} } \times \frac{3 + \sqrt{2} }{3 + \sqrt{2} } \\ \\ = \frac{(3 + \sqrt{2}) {}^{2} }{(3) {}^{2} - ( \sqrt{2} ) {}^{2} } \\ \\ = \frac{9 + 2 + 6 \sqrt{2} }{9 - 2} \\ \\ \frac{11 + 6 \sqrt{2} }{7} = a + b \sqrt{2} \\ \\ = a = \frac{11}{7} \\ \\ b = \frac{6 }{11}$

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2) Please refer the attachment

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$3)x = 3 + 2 \sqrt{2} \\ \\ \frac{1}{x} = \frac{1}{3 + 2 \sqrt{2} } \times \frac{3 - 2 \sqrt{2} }{3 - 2 \sqrt{2} } \\ \\ = \frac{3 + 2 \sqrt{2} }{(3) {}^{2} - (2 \sqrt{2}) {}^{2} } \\ \\ = \frac{3 - 2 \sqrt{2} }{9 - 8} \\ \\ = 3 - 2 \sqrt{2}$

$∴x + \frac{1}{x} = 3 + 2 \sqrt{2} + 3 - 2 \sqrt{2} \\ \\ x + \frac{1}{x} = 3 + 3 = 6 \\ \\ x + \frac{1}{x} = 6 - 2 = 4 \\ \\ = (\sqrt{x} - \frac{ 1 }{ \sqrt{x} } ) {}^{2} = 4 \\ \\ \sqrt{x} - \frac{1}{ \sqrt{x} } = \sqrt{4} \\ \\ \sqrt{x} - \frac{1}{ \sqrt{x} } = 2$

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