Question

please answer this question fast tomorrow is my exam​

Answer 1

Answer:

14

Step-by-step explanation:

$Given: x=\frac{2+\sqrt{3}}{2-\sqrt{3}}$

$=\frac{2+\sqrt{3}}{2-\sqrt{3}}*\frac{2+\sqrt{3}}{2+\sqrt{3}}$

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

$=\frac{4+3+4\sqrt{3}}{4-3}$

$=\frac{7+4\sqrt{3}}{1}$

$=7 + 4\sqrt{3}$

Then,

$=>\frac{1}{x} = \frac{1}{7+4\sqrt{3}} * \frac{7-4\sqrt{3}}{7-4\sqrt{3}} = \frac{7-4\sqrt{3}}{(7)^2 - (4\sqrt{3})^2} = \frac{7-4\sqrt{3}}{49-48} = 7-4\sqrt{3}$

Now,

x + (1/x) = 7 + 4√3 + 7 - 4√3

           = 14.

Hope it helps!

Answer 2

Answer:

1/x=2-root3/2+root3 put in x+1/x

2+root3/2-root3 + 2-root3/2+root3=

solve it- (2+root3) ^2 + (2-root3)^2 / (2-root3)(2+root3)

8+6/4-3=14/1=14ans.