Question

Solve the problem plzzzzzzzzzzz​

Answer 1

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$\sqrt{x} = \sqrt{3 + 2 \sqrt{2} } = \sqrt{p} + \sqrt{q} \\ \\ since \\ \: \: 3 + 2 \sqrt{2} = p + q + 2 \sqrt{pq} \\ = = > p + q = 3 \\ pq = 2 \\ \\ since \\ p - q = 1 \\ = = > p = 2 \\ q = 1 \\ \\ since \\ \sqrt{x} = \sqrt{2} + \sqrt{1} = 1 + \sqrt{2} \\ \\ since \: \: \sqrt{x} - \frac{1}{ \sqrt{x} } =( 1 + \sqrt{2} ) - \frac{1}{1 + \sqrt{2} } \\ \\ = \frac{(1 + \sqrt{ {2}^{2} } - 1 }{1 + \sqrt{2} } \\ \\ = \frac{1 + 2 + 2 \sqrt{2} }{1 + \sqrt{2} } \\ \\ = \frac{(2(1 + \sqrt{2)} }{1 + \sqrt{2} } \\ \\ = = > 2$

Answer 2

ANSWER

Plz refer to the attachment for detailed answer here.

Explanation

1)Put the value of x on the given expression and then and then simply them.

2)Use componendo and dividendo later to rationalise the denominator.

3)Also remember the algebraic expressions.

(a+b)^2 = a^2 + b^2 + 2ab

a^2 - b^2 = (a+b)(a-b)