Question

Someone pls answer this question, it's really important, you'll get 20 points. and I'll mark you brainliest​

Answer 1

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

Now rationalize the denominator . We get ,

$\implies \bf \dfrac{3+\sqrt{6}}{\sqrt{3}+\sqrt{2}}*\dfrac{\sqrt{3}-\sqrt{2} }{\sqrt{3}-\sqrt{2}}=a+\sqrt{3}b\\\\ \implies \bf \dfrac{3\sqrt{3}-3\sqrt{2}+\sqrt{18}-\sqrt{12} }{(\sqrt{3})^2-(\sqrt{2})^2}=a+\sqrt{3}b\\\\ \implies \bf \dfrac{3\sqrt{3}-3\sqrt{2}+3\sqrt{3}-2\sqrt{3} }{1}=a+\sqrt{3}b\\\\ \implies \bf -3\sqrt{2}+\sqrt{3}(-2)=a+\sqrt{3}b$

Now compare both LHS & RHS , we get ,

$\implies \bold{\sf {\orange{a=-3\sqrt{2}\;\;\;\&\;\;\;b=-2}}}$