Question

Do it for me the 2 one (ls-number system) do it in a proper manner air u can say step by step
Best answer will be marked as brainliest

Answer 1

Step-by-step explanation:

Hey mate

$\huge\underline\bold{Solution}$

$x = \sqrt{3} - 2$

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

Now rationalise it

$\frac{1}{x} = \frac{1}{ \sqrt{3} - 2 } \times \frac{ \sqrt{3} + 2 }{ \sqrt{3} + 2}$

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

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

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

To find

$x - \frac{1}{x}$

put the value of x and 1/x

$( \sqrt{3} - 2) - ( - \sqrt{3} - 2)$

$\sqrt{3} - 2 + \sqrt{3} + 2$

$\sqrt{3} + \sqrt{3}$

$2 \sqrt{3}$

Now again to find

${x}^{2} + \frac{1}{ {x}^{2} }$

Put the value of x and 1/x

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

$(3 + 4 - 2 \times \sqrt{3} \times 2) + ( - (3 + 4 + 2 \times \sqrt{3} \times 2))$

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

$- 8 \sqrt{3}$

Hope this will help u

And plzzz

$\huge\underline\red{BRAINLIST}$$\huge\blue{PLEASE}$