Question

if x=2+√3 then x+1 by x is​

Answer 1

Step-by-step explanation:

$if \: \: x=2+ \sqrt{3}$

Find :-

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

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

On rationalise the denominator :-

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

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

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

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

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

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

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

$= 4$

I hope it is helpful

Answer 2

given

x= 2 + √3

so we can write

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

rationalizing the denominator we get

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

Now

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