Discussion

Explanation:

Rationalizing the first term here, $\frac{2}{\sqrt{3}-\sqrt{2}}$ = $\frac{2}{\sqrt{3}-\sqrt{2}}\times \frac{\sqrt{3}+\sqrt{2}}{\sqrt{3}+\sqrt{2}}$ = 2√3 + 2√2.

Rationalizing the second term here, $\frac{1}{3-2\sqrt{2}}$ = $\frac{1}{3-2\sqrt{2}}\times \frac{3+2\sqrt{2}}{3+2\sqrt{2}}$ = 3 + 2√2

∴ $\frac{2}{\sqrt{3}+\sqrt{2}}-\frac{1}{3-2\sqrt{2}}$ = 2√3 + 2√2 – (3 + 2√2) = 2√3 – 3 = √3(2 - √3)

Hence, option (b).

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