Question

$\sqrt{3} - 1 \div \sqrt{3 } + 1 = a + b \sqrt{3}$
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Answer 1

Step-by-step explanation:

▪️√3-1\√3+1=a-b√3

▪️(Rationalize the denominator)

▪️In LHS(left hand side)

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)

(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)

2-√3=a-b√3 (Comparing both equations)So, a=2,b=1.

Hopes it help you❤️❤️

Answer 2

Step-by-step explanation:

$\frac{ \sqrt{3} - 1 }{ \sqrt{3} + 1} =( a + b) \sqrt{3} \\ rationalise \: the \: denominator \\ \frac{ {( \sqrt{3} - 1)}^{2} }{3 - 1} = (a + b) \sqrt{3} \\ \fbox{ a + b = \frac{ ({ \sqrt{3} - 1})^{2} }{2 \sqrt{3} }}$