Question

please solve..........................................................iodjoi[d.d.

Answer 1

(√3-1)/(√3+1)
=[(√3-1)/(√3+1)]×[(√3-1)/(√3-1)]

=(3+1-2√3)/(3-1)

=(4-2√3)/2

=[2(2-√3)]/2

=2-√3
a-b√3=2-√3
∴the values of a=2 and b=1
So the values of a and b are 2 and 1 respectively.



Hope it helps...
Please mark my answer as the brainliest...

Answer 2

$\displaystyle\huge\red{\underline{\underline{Solution}}}$

FORMULA TO BE IMPLEMENTED

$\green{{if \: x + y \sqrt{3} = a + b \sqrt{3} \: \: \: then \: \: x = a , y = b}}$

TO DETERMINE

There value of a & b where

$\displaystyle \: \frac{ \sqrt{ {3} } - 1 }{ \sqrt{3} + 1 } = a + b \sqrt{3}$

PROCESS

In order to find the required value

• we have multiply numerator and denominator if LHS both by the conjugate surd of denominator

• Then to compare both sides

CALCULATION

$\displaystyle \: \frac{ \sqrt{ {3} } - 1 }{ \sqrt{3} + 1 } = a + b \sqrt{3}$

$\implies \: \displaystyle \: \frac{( \sqrt{ {3} } - 1)( \sqrt{3} - 1) }{ (\sqrt{3} + 1)( \sqrt{3} - 1) } = a + b \sqrt{3}$

$\implies \: \displaystyle \: \frac{{( \sqrt{ {3} } - 1)}^{2} }{ {( \sqrt{3})}^{2} - {(1)}^{2} } = a + b \sqrt{3}$

$\implies \: \displaystyle \: \frac{{( { {3} } - 2 \sqrt{3} + 1)}}{ 3 - 1} = a + b \sqrt{3}$

$\implies \: \displaystyle \: \frac{{4 - 2\sqrt{ {3} } } }{2} = a + b \sqrt{3}$

$\implies \: \displaystyle \: \frac{{2(2 - \sqrt{ {3} } } )}{2} = a + b \sqrt{3}$

$\implies \: \displaystyle \: 2 - \sqrt {3} = a + b \sqrt{3}$

$\implies \: \displaystyle \: a + b \sqrt{3} = 2 - \sqrt{3}$

$\implies \: \displaystyle \: a \: = 2 \: \: \: \: and \: \: b \: = - 1$

RESULT

$\red{\fbox {a = 2 \: \: , b = - 1\: } \: }$