Question

if x 1/ √ 3- √ 2then find the value of (x²+1/x²)
please help me guys ​

Answer 1

Answer:

$x = \frac{1}{ \sqrt{3} - \sqrt{2} } \\ x= \frac{1}{ \sqrt{3} - \sqrt{2} } \times \frac{ \sqrt{3 } + \sqrt{2} }{ \sqrt{3} + \sqrt{2} } \\ x = \frac{ \sqrt{3} + \sqrt{2} }{ { \sqrt{3} }^{2} - { \sqrt{2} }^{2} } \\ x = \frac{ \sqrt{3} + \sqrt{2} }{1} \\ \frac{1}{x} = \frac{1}{ \frac{1}{ \sqrt{3} - \sqrt{2} } } \\ \frac{1}{x} = \sqrt{3} - \sqrt{2} \\ {x}^{2} + \frac{1}{ {x}^{2} } = {( \sqrt{3} + \sqrt{2}) }^{2} + {( \sqrt{3} - \sqrt{2)} }^{2} \\ = { \sqrt{3} }^{2} + { \sqrt{2} }^{2} + 2 \sqrt{3} \sqrt{2} + { \sqrt{3} }^{2} + { \sqrt{2} }^{2} - 2 \sqrt{3} \sqrt{2} \\ = 3 + 2 + 2 \sqrt{6} + 3 + 2 - 2 \sqrt{6} \\ = 5 + 5 \\ = 10$

Answer 2

$x = \frac{1}{ \sqrt{3} - \sqrt{2} } \\ x = \frac{ \sqrt{3} + \sqrt{2} }{3 - 2} \\ x = \sqrt{3} + \sqrt{2} \\ \\ \frac{1}{x} = \sqrt{3} - \sqrt{2} \\ \\ x + \frac{1}{x} = \sqrt{3} + \sqrt{2} + \sqrt{3} - \sqrt{2} \\ x + \frac{1}{x} = 2 \sqrt{3} \\ ( {x + \frac{1}{x} })^{2} = 10 \\ {x}^{2} + \frac{1}{ {x}^{2} } + 2 = 10 \\ {x}^{2} + \frac{1}{ {x}^{2} } = 10$