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Question

Simplify by rationalizing the denominator $$\displaystyle \dfrac{2\sqrt{3}-\sqrt{2}}{2\sqrt{2}+3\sqrt{3}}$$


A
207619
loader
B
227619
loader
C
217619
loader
D
22769
loader

Solution

The correct option is B $$\displaystyle \dfrac{22-7\sqrt{6}}{19}$$
$$\displaystyle \frac{2\sqrt{3}-\sqrt{2}}{2\sqrt{2}+3\sqrt{3}}$$
the conjugate of $$2 \sqrt{2}+3 \sqrt{3}$$ is $$2\sqrt{2}-3\sqrt{3}$$
$$=\displaystyle \frac{2\sqrt{3}-\sqrt{2}}{2\sqrt{2}+3\sqrt{3}} \times$$ $$\displaystyle \frac{2\sqrt{2}-3\sqrt{3}}{2\sqrt{2}-3\sqrt{3}}$$
$$=\displaystyle \frac{4\sqrt{6}-6\sqrt{9}-2 \sqrt{4}+3 \sqrt{6}}{8-27}$$
$$=\displaystyle \frac{4 \sqrt{6}-18-4+3 \sqrt{6}}{-19}=\frac{7\sqrt{6}-22}{-19}$$
$$=\displaystyle \frac{22-7\sqrt{6}}{19}$$

Mathematics

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