Question

√2+√3/3√2-2√3 find b
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Answer 1

Answer:

Step-by-step explanation:

I Hope You Understand ;-)

Answer 2

Correct Question -

$\sf{ \dfrac{ \sqrt{2} + \sqrt{3} }{ 3 \sqrt{2} - 2 \sqrt{3} } = 2a - b \sqrt{6} }$

SoLuTiOn -

$</p><p>\sf{ \dfrac{ \sqrt{2} + \sqrt{3} }{ 3 \sqrt{2} - 2 \sqrt{3} } = 2a - b \sqrt{6} } \\ \\ \sf{ => \dfrac{ \sqrt{2} + \sqrt{3} }{ 3 \sqrt{2} - 2 \sqrt{3} } } \: \times \dfrac{ 3 \sqrt{2} + 2 \sqrt{3} }{ 3 \sqrt{2} + 2 \sqrt{3} } = 2a - b \sqrt{6} \: \\ \\ \dfrac{(2 + \sqrt{3} )\times (3 \sqrt{2} + 2 \sqrt{3} ) }{6} = 2a - b \sqrt{6} \\ \\ \\ 6 \sqrt{2 }+ 4 \sqrt{3} + 3 \sqrt{6} + 6 = 2a - b \sqrt{6} \\ \\ (6 \sqrt{2} + 4 \sqrt{3} + 6) + (3 \sqrt{6 } ) = 2a - b \sqrt{6} \\ \\ \sf{ \: comparing \: we \: get \: - } \: \\ \\ \sf{ \: a \: =3 \sqrt{2} + 2 \sqrt{3} + 3 } \: and \: b \: = - 3$