Question

√3 + 1/ √3 - 1 = a + b √3​

Answer 1

Answer:

√3-1\√3+1=a-b√3

(Rationalize the denominator)

In LHS(left hand side)

= (√3-1)(√3-1)\(√3+1)(√3-1)

= 3+1-2√3\3-1

=4-2√3\2

(Taking 2 common from numerator)

=2-√3 (1)

In RHS

a-b√3. (2)

A.T.Q(according to question)

(1)=(2)

2-√3=a-b√3 (Comparing both equations)

So, a=2,b=1.

Step-by-step explanation:

follow me

Answer 2

Answer:

According to your question we have to find the value of A and B by solving the equation given above

First of all we have to rationalize the denominator then we will be able to find the value of A and B

let us find

$\frac{ \sqrt{3} + 1 }{ \sqrt{3} - 1 } = a + b \sqrt{3} \\ \\ = \geqslant \frac{ \sqrt{3 } + 1}{ \sqrt{3} - 1} \times \frac{ \sqrt{3} + 1 }{ \sqrt{3} + 1 } = a + b \sqrt{3} \\ \\ = \geqslant \frac{( \sqrt{3} + 1) {}^{2} }{( \sqrt{3}) {}^{2} - {(1)}^{2} } = a + b \sqrt{3} \\ \\ = \geqslant \frac{3 + 1 + 2\sqrt{3} }{3 - 1} = a + b \sqrt{3} \\ \\ = \geqslant \frac{4 + 2 \sqrt{3} }{2} = a + b \sqrt{3} \\ \\ = \geqslant \frac{2(2 + \sqrt{3}) }{2} = a + b \sqrt{3} \\ \\ = \geqslant 2 + \sqrt{3} = a + b \sqrt{3} \\ \\ = \geqslant a = 2 \\ \\ = \geqslant b \sqrt{3} = \sqrt{3} = 1 \\ \\ \sqrt{3} \: will \: be \: cancelled \: from \: \sqrt{3}$