Question

a=3+2√2 and ab=1 then value of a2+3ab+b2÷a2-3ab+b2​

Answer 1

$a = 3 + 2 \sqrt{2} \\ {a }^{2} = 9 + 8 + 12 \sqrt{2}$

given ab=1

=》b=1/a

$b = \frac{1}{3 + 2 \sqrt{2} } \\ b = \frac{3 - 2 \sqrt{2} }{(3 + 2 \sqrt{2})(3 - 2 \sqrt{2}) } \\ b = \frac{3 - 2 \sqrt{2} }{9 - 8} \\ b = 3 - 2 \sqrt{2} \\ {b}^{2} = 9 + 8 - 12 \sqrt{2}$

a^2+3ab+b^2÷a^2-3ab+b^2

=(a^2+b^2+3)/(a^2+b^2-3)

$\frac{ {a}^{2} + {b}^{2} + 3 }{ {a}^{2} + {b}^{2} - 3 } = \frac{(17 + 12 \sqrt{2} ) + (17 - 12 \sqrt{2} ) + 3}{(17 + 12 \sqrt{2}) + (17 - 12 \sqrt{2} ) - 3 } = \frac{37}{31}$

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Answer 2

Answer:

given

$a = 3 + 2 \sqrt{2} \\ \\ ab = 1$

Now

$b = \frac{1}{a} \\ \\ = \frac{1}{3 + 2 \sqrt{2} } \\ \\ = \frac{3 - 2 \sqrt{2} }{9 - 8} \\ \\ = 3 - 2 \sqrt{2}$

$\frac{ {a}^{2} + 3ab + {b}^{2} }{ {a}^{2} - 3ab + {b}^{2} } \\ \\ = \frac{ {a}^{2} + {b}^{2} + 3}{ {a }^{2} + {b}^{2} - 3} \\ \\ = \frac{( {a + b})^{2} - 2 + 3 }{( {a + b})^{2} - 2 - 3} \\ \\ = \frac{ {6}^{2} + 1 }{ {6}^{2} - 5} \\ \\ = \frac{37}{31}$