Explanation


Let’s translate this question, one step at a time.

a= \(3\sqrt{2}\), b = \(2\sqrt{3}\) and c= \(2\sqrt{6}\)

x = \((3\sqrt{2}+ 2\sqrt{3})^2\)
= \((3\sqrt{2}+ 2\sqrt{3})(3\sqrt{2}+ 2\sqrt{3})\)

= \(30+12\sqrt{6}\)

y = \(6(5 - 2\sqrt{6})\)=\(30-12\sqrt{6}\)

z = \(2^2*3^2\) =\(36\).

\(\frac{xy}{z}\)=\((30-12\sqrt{6})(30+12\sqrt{6})\frac{1}{36}\) = 1

Hence option A is correct.

Did you recognize the common quadratics? The product of x and y in the last chunk of the question contains one that could save you that intermediary step: \((x + y)(x - y) = x^2 - y^2\). Of course, it doesn’t take too much longer to write out the math.
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You said x = 3 + 12*sqrt(6). This is not correct.

x = (3*sqrt(2)+2*sqrt(3))^2 = 30 + 12*sqrt(6).

Thanks for pointing out!
It was a typo
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