Question

answer fast please please

Answer 1

Solution :

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Given :

To find a & b,

where they are rational number in,

⇒ $\frac{ \sqrt{2} + \sqrt{3} }{3 \sqrt{2} - 2\sqrt{3} } = a - b \sqrt{3}$


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By taking conjugate,

We get,.

⇒ $\frac{ \sqrt{2} + \sqrt{3} }{3 \sqrt{2} - 2\sqrt{3} } ( \frac{3 \sqrt{2} + 2 \sqrt{3} }{3 \sqrt{2} + 2 \sqrt{3}}) = a - b\sqrt{3}$

⇒ $\frac{(\sqrt{2} + \sqrt{3})(3 \sqrt{2} + 2\sqrt{3}) }{(3 \sqrt{2} - 2 \sqrt{3} )(3 \sqrt{2} + 2 \sqrt{3})} = a - b\sqrt{3}$

⇒ $\frac{ \sqrt{2}(3 \sqrt{2} ) + \sqrt{2}(2 \sqrt{3} ) + \sqrt{3}(3 \sqrt{2} ) + \sqrt{3} (2 \sqrt{3} ) }{(3 \sqrt{2} )^2 - (2 \sqrt{3})^2} = a - b\sqrt{3}$

⇒ $\frac{ 6 + 2 \sqrt{6} + 3 \sqrt{6} + 6 ) }{18 - 12 } = a - b\sqrt{3}$

⇒ $\frac{12 + 5 \sqrt{6} }{6}  = a - b\sqrt{3}$

⇒ $\frac{12}{6} + \frac{5 \sqrt{6} }{6}  = a - b\sqrt{3}$

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

⇒ $2 + \frac{5}{ \sqrt{6} }  = a - b\sqrt{3}$

∴ a - b√3 = $2 + \frac{5}{ \sqrt{6} }$

⇒ ∴ a = 2

&

⇒ b√3 =  $\frac{5}{ \sqrt{6} }$

⇒ b = $\frac{5}{ \sqrt{6} }( \frac{1}{ \sqrt{3} } )$

⇒ b = $\frac{5}{ \sqrt{18} }$

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                                        Hope it Helps !!

⇒ Mark as Brainliest,.

Answer 2

heya....

i think something is mistake in your questions

right questions is like that:-

find a and b where they are irrational no. in
$\frac{ \sqrt{2} + \sqrt{3} }{3 \sqrt{2} - 2 \sqrt{3} } = a - b \sqrt{6}$




$\frac{ \sqrt{2} + \sqrt{3} }{3 \sqrt{2} - 2 \sqrt{3} } \\ \\ = > \frac{( \sqrt{2} + \sqrt{3} )}{3 \sqrt{2} - 2 \sqrt{3} } \times \frac{3 \sqrt{2} + 2 \sqrt{3} }{3 \sqrt{2} + 2 \sqrt{3} } \\ \\ = > \frac{6 + 2 \sqrt{6} +3 \sqrt{6} + 6 }{18 - 12} \\ \\ = > \frac{12 + 5 \sqrt{6} }{6} \\ \\ = > \frac{12}{6} + \frac{5 \sqrt{6} }{6} \\ \\ = > 2 + \frac{5}{ \sqrt{6} }$


NOW,

on comparing L.H.S = R.H.S

a = 2 ,b = -5/√6