Question

please answer this question fastly

Answer 1

Hello Mate!

$a = \frac{ \sqrt{3} - \sqrt{2} }{ \sqrt{3} + \sqrt{2} } \times \frac{ \sqrt{3} - \sqrt{2} }{ \sqrt{3} - \sqrt{2} } \\ \frac{ {( \sqrt{3} - \sqrt{2} )}^{2} }{3 - 2} = \\ 3+ 2 - 2 \sqrt{6} \\ 5 - 2 \sqrt{6}$

$b = \frac{ \sqrt{3} + \sqrt{2} }{ \sqrt{3} - \sqrt{2} } \times \frac{ \sqrt{3} + \sqrt{2} }{ \sqrt{3} + \sqrt{2} } \\ = \frac{ { (\sqrt{3} + \sqrt{2} ) }^{2} }{3 - 2} \\ = 5 + 2 \sqrt{6}$
${a}^{2} = {(5 - 2 \sqrt{6} )}^{2} \\ 25 + 4 \times 6 - 44 \sqrt{6} \\ 49 - 44 \sqrt{6} \\ {b}^{2} = {(5 + 2 \sqrt{6}) }^{2} \\ 25 + 24 + 44 \sqrt{6} \\ 49 + 44 \sqrt{6}$
${a}^{2} + {b}^{2} \\ 49 - 44 \sqrt{6} + 49 + 44 \sqrt{6} \\ 98$

Hope it helps☺!

Answer 2

first rationalize the value of a and b:-
a=√3-√2÷√3+√2×√3-√2÷√3-√2
=> (√3+√2)²÷(√3)²-(√2)²
=>3-2+2√6÷3-2
=>1+6√2=> value of a

lly , rationalize the value of b:-

b=√3+√2÷√3-√2
=>√3-√2÷√3-√2×√3+√2÷√3+√2
=>(√3+√2)²÷(√3)²-(√2)²
=>3+2+2√6÷3-2
=>6+6√2=> value of b

now put this two values in the formula a²-b²

=> a²-b²=(1+6√2)²-(6+6√2)²
=>(1+72)-(36+72)
=>73-108
=>-35 Answer

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