Question

$22 \div 2 \sqrt{3 } - 1 + 17 \div 2 \sqrt{3} - 1$
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Answer 1

Answer:

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Answer 2

22 ÷ 2$\sqrt{3}-$ 1 + 17 ÷ 2$\sqrt{3}$- 1=$\dfrac{(13\sqrt{3} -4)}{2}$

Step-by-step explanation:

We have,

22 ÷ 2$\sqrt{3}$ - 1 + 17 ÷ 2$\sqrt{3}$ - 1

To find, the value of 22 ÷ 2$\sqrt{3}$ - 1 + 17 ÷ 2$\sqrt{3}$ - 1=?

∴ 22 ÷ 2$\sqrt{3}$ - 1 + 17 ÷ 2$\sqrt{3}$ - 1

$=\dfrac{22}{2\sqrt{3}} -1+\dfrac{17}{2\sqrt{3}}-1$

$=\dfrac{22}{2\sqrt{3}} +\dfrac{17}{2\sqrt{3}}-1-1$

$=\dfrac{22+17}{2\sqrt{3}} -2$

$=\dfrac{39}{2\sqrt{3}} -2$

$=\dfrac{39-4\sqrt{3}}{2\sqrt{3}}$

$=\dfrac{(13\sqrt{3} -4)\sqrt{3}}{2\sqrt{3}}$

$=\dfrac{(13\sqrt{3} -4)}{2}$

∴ 22 ÷ 2$\sqrt{3}$- 1 + 17 ÷ 2$\sqrt{3}$- 1=$\dfrac{(13\sqrt{3} -4)}{2}$

Hence, 22 ÷ 2$\sqrt{3}-$ 1 + 17 ÷ 2$\sqrt{3}$- 1=$\dfrac{(13\sqrt{3} -4)}{2}$