Question

if a=√3+√2/√3-√2 and b=√3-√2/√3+√2
then value of a+b is ​

Answer 1

Answer:

a+b=10√6

Step-by-step explanation:

a=√3+√2/√3-√2

b=√3-√2/√3+√2

a+b=(√3+√2)²+(√3-√2)²/√3²-√2²

=3+2√6+2+3-2√6+2√6/3-2

10√6/1

∴the answer is 10√6

Hope this might help you....

Answer 2

Answer:

Given, a = √ 3 + √2 / √3 - √2

and

b = √ 3 - √ 2 / √3 + √ 2

so putting back the value, we get :

$a + b \\ \\ = \frac{ \sqrt{3} + \sqrt{2} }{ \sqrt{3} - \sqrt{2} } + \frac{ \sqrt{3} - \sqrt{2} }{ \sqrt{3} + \sqrt{2} } \\ \\ = \frac{( \sqrt{3} + \sqrt{2}) ^{2} + ( \sqrt{3} - \sqrt{2} ) ^{2} }{( \sqrt{3} - \sqrt{2} )( \sqrt{3} + \sqrt{2}) } \\ \\ = \frac{ (\sqrt{3})^{2} + 2 \times \sqrt{3 } \times \sqrt{2} + ( \sqrt{2} ) ^{2} + ( \sqrt{3} - \sqrt{2}) ^{2} }{( \sqrt{3}^{2} - ( \sqrt{2})^{2} } \\ \\ = \frac{3 + 2 \sqrt{6} + 2 + ( { \sqrt{3} })^{2} - 2 \times \sqrt{3} \times \sqrt{2} + ( \sqrt{2} )^{2} }{3 - 2} \\ \\ = \frac{5 + 2 \sqrt{6} + 3 - 2 \sqrt{6} + 2 }{1} \\ \\ = \frac{5 + 3 + 2}{1} \\ \\ ( + 2 \sqrt{6} \: and \: - 2 \sqrt{6} \: cancelled) \\ \\ = \frac{10}{1} \\ \\ = 10 \\ \\ formula \: used \\ \\ (a + b) ^{2} = a ^{2} + 2ab + b ^{2} \\ \\ (a - b) ^{2} = a ^{2} - 2 ab + b ^{2} \\ \\ a ^{2} - b ^{2} = ( a + b)(a - b)$

↔ a + b = 10 .

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hope it helps uH ✔