If frac{2 sqrt{6}-sqrt{5}}{3 sqrt{5}-2 sqrt{6 | vedclass

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(A) Given: $\frac{2 \sqrt{6}-\sqrt{5}}{3 \sqrt{5}-2 \sqrt{6}}=a+b \sqrt{30}$To rationalize the denominator,multiply the numerator and denominator by $

Vedclass · Accepted Answer

$a=\frac{3}{7}, b=\frac{4}{21}$

Vedclass · Answer

(A) Given: $\frac{2 \sqrt{6}-\sqrt{5}}{3 \sqrt{5}-2 \sqrt{6}}=a+b \sqrt{30}$To rationalize the denominator,multiply the numerator and denominator by $(3 \sqrt{5}+2 \sqrt{6})$:$\frac{2 \sqrt{6}-\sqrt{5}}{3 \sqrt{5}-2 \sqrt{6}} 	imes \frac{3 \sqrt{5}+2 \sqrt{6}}{3 \sqrt{5}+2 \sqrt{6}}=a+b \sqrt{30}$Using the identity $(x-y)(x+y) = x^2 - y^2$ in the denominator:$\frac{(2 \sqrt{6}-\sqrt{5})(3 \sqrt{5}+2 \sqrt{6})}{(3 \sqrt{5})^{2}-(2 \sqrt{6})^{2}}=a+b \sqrt{30}$Expanding the numerator:$(2 \sqrt{6})(3 \sqrt{5}) + (2 \sqrt{6})(2 \sqrt{6}) - (\sqrt{5})(3 \sqrt{5}) - (\sqrt{5})(2 \sqrt{6}) = 6 \sqrt{30} + 24 - 15 - 2 \sqrt{30} = 9 + 4 \sqrt{30}$Calculating the denominator:$(3 \sqrt{5})^2 - (2 \sqrt{6})^2 = 45 - 24 = 21$So,$\frac{9+4 \sqrt{30}}{21} = a+b \sqrt{30}$$\frac{9}{21} + \frac{4 \sqrt{30}}{21} = a+b \sqrt{30}$$\frac{3}{7} + \frac{4}{21} \sqrt{30} = a+b \sqrt{30}$Comparing the rational and irrational parts,we get $a=\frac{3}{7}$ and $b=\frac{4}{21}$.

If $\frac{2 \sqrt{6}-\sqrt{5}}{3 \sqrt{5}-2 \sqrt{6}}=a+b \sqrt{30}$,find the value of $a$ and $b$.

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