Question

plzz zzz solve this problem
Ques 23,24,25

Answer 1

heya....

Here is your answer....

Answer 2

Here is your solution

23.
$x = 3 + 2 \sqrt{2} \\ \\ \frac{1}{x} = \frac{1}{3 + 2 \sqrt{2} } \\ \frac{1}{x} = \frac{1}{3 + 2 \sqrt{2} } \times \frac{3 - 2 \sqrt{2} }{3 - 2 \sqrt{2} } \\ \\ \frac{1}{x} = \frac{3 - 2 \sqrt{2} }{(3 ){}^{2} - (2 \sqrt{2} ){}^{2} } \\ \\ \frac{1}{x} = \frac{3 - 2 \sqrt{2} }{9 - 8} \\ \frac{1}{x} = 3 - 2 \sqrt{2} \\ \\ x + \frac{1}{x} \\ x + \frac{1}{x} = 3 + 2 \sqrt{2} + 3 - 2 \sqrt{2} \\ x + \frac{1}{x} = 6$

25.
$\frac{4 + \sqrt{5} }{4 - \sqrt{5} } \times \frac{4 - \sqrt{5} }{4 + \sqrt{5} } \\ \\ = \frac{(4 ){}^{2} - ( \sqrt{5} ) {}^{2} }{(4) {}^{2} - (\sqrt{5} ){}^{2} } \\ \\ = \frac{16 - 5}{16 - 5} \\ = \frac{11}{11} \\ = 1$