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Question

Find the value of $$\dfrac{\sqrt{32}+\sqrt{48}}{\sqrt8+\sqrt{12}}.$$


A
22+3
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B
2+3
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C
2
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D
1
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Solution

The correct option is C $${2}$$

$$\displaystyle(\frac{\sqrt {32}+\sqrt {48}}{\sqrt {8}+\sqrt{12}})\\= (\dfrac{\sqrt {16\times 2}+\sqrt {16\times 3}}{\sqrt {4\times 2}+\sqrt{4\times 3}})\\(\dfrac{4\sqrt {2}+4\sqrt {3}}{2\sqrt {2}+2\sqrt{3}})= (\dfrac{4}{2})(\dfrac{\sqrt {2}+\sqrt {3}}{\sqrt {2}+\sqrt{3}})=2$$


Mathematics

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