Question

If x=2-√3 then the value of x × x +1÷x × x and x × x - 1 ÷ x × x is

Answer 1

My understanding of the question is this:

If $x = 2- \sqrt{3}$  then find the value of $x^2 + \frac{1}{x^2 }$ and $x^2 - \frac{1}{x^2 }$

Solution:
$x = 2- \sqrt{3}$
Squaring on both sides

$x^2 = 4 + 3 - 4 \sqrt{3} = 7-4 \sqrt{3}$

$\frac{1}{x^2} = \frac{1}{7-4 \sqrt{3}}$

On Rationalizing

$\frac{1}{x^2} = \frac{1}{7-4 \sqrt{3}}. \frac{7+4 \sqrt{3}}{7+4 \sqrt{3}}$

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

Now, 

$x^2 + \frac{1}{x^2 } = 7-4 \sqrt{3} + 7+ 4 \sqrt{3} = 14$

And,

$x^2 - \frac{1}{x^2 } = 7-4 \sqrt{3} - 7- 4 \sqrt{3} = -8 \sqrt{3}$