Question

the value of 4root3-4/root3+1 - 2-root3/2+root3 is. pls tell me quick​

Answer 1

$} [/t</p><p>[tex] \frac{4 \sqrt{3} - 4 }{ \sqrt{3} + 1} - \frac{2 - \sqrt{3} }{2 + \sqrt{3} } = \frac{8 \sqrt{3} - 8 + 4 \times 3 - 4 \sqrt{3} - (2 \sqrt{3} - 3 + 2 - \sqrt{3)} }{ (\sqrt{3} + 1) \times (2 + \sqrt{3)} } = \frac{8 \sqrt{3} - 8 + 12 - 4 \sqrt{3} - \sqrt{3} + 1}{2 \sqrt{3} + 3 + 2 + \sqrt{3} } = \frac{3 \sqrt{3 } + 5}{3 \sqrt{3} + 5} = 1$

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Answer 2

Answer:

The value is 1

Explanation:

Given: The $\frac{4\sqrt{3}-4}{\sqrt{3}+1}-\frac{2-\sqrt{3}}{2+\sqrt{3}}$

Find: The value of given equation.

Solution:

$\frac{4\sqrt{3}-4}{\sqrt{3}+1}-\frac{2-\sqrt{3}}{2+\sqrt{3}}$

$\frac{(4\sqrt{3}-4)(2+\sqrt{3})-(2-\sqrt{3})(\sqrt{3}+1)}{(\sqrt{3}+1)(2+\sqrt{3})}$

$\frac{8\sqrt{3}+12-8-4\sqrt{3}-(2\sqrt{3}+2-3-\sqrt{3})}{2\sqrt{3}+3+2+\sqrt{3}}$

$\frac{3\sqrt{3}+5}{3\sqrt{3}+5}=1$

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