Question

2+√3/2-√3+2-√3/2+√3+√3-1/√3+1 simplify​

Answer 1

Answer: 16-√3

Step-by-step explanation:

In attachment.

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Answer 2

Answer:

$16 - \sqrt{3}$ is the required value of $\frac{2 + \sqrt{3} }{2 - \sqrt{3} } + \frac{2 - \sqrt{3}}{2 + \sqrt{3} } + \frac{\sqrt{3}-1 }{\sqrt{3}+1 }$.

Step-by-step explanation:

Explanation:

Given in the question,  $\frac{2 + \sqrt{3} }{2 - \sqrt{3} } + \frac{2 - \sqrt{3}}{2 + \sqrt{3} } + \frac{\sqrt{3}-1 }{\sqrt{3}+1 }$.

Rationalization - An irrational number is rationalized, or changed into a number that can be stated as the ratio of two integers, through the process of rationalization.

We first rationalize the given numbers,

⇒ $[\frac{2 + \sqrt{3} }{2 - \sqrt{3} } .\frac{2 + \sqrt{3} }{2 + \sqrt{3} } ]+ [\frac{2 - \sqrt{3}}{2 + \sqrt{3} } . \frac{2 - \sqrt{3} }{2 - \sqrt{3} } ] + [\frac{\sqrt{3}-1 }{\sqrt{3}+1 }.\frac{\sqrt{3} -1}{\sqrt{3} -1} ]$

Step 1:

We have, $\frac{2 + \sqrt{3} }{2 - \sqrt{3} } + \frac{2 - \sqrt{3}}{2 + \sqrt{3} } + \frac{\sqrt{3}-1 }{\sqrt{3}+1 }$.

On rationalizing we get,

$[\frac{2 + \sqrt{3} }{2 - \sqrt{3} } .\frac{2 + \sqrt{3} }{2 + \sqrt{3} } ]+ [\frac{2 - \sqrt{3}}{2 + \sqrt{3} } . \frac{2 - \sqrt{3} }{2 - \sqrt{3} } ] + [\frac{\sqrt{3}-1 }{\sqrt{3}+1 }.\frac{\sqrt{3} -1}{\sqrt{3} -1} ]$

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

[where $a^2 -b^2 = (a+b)(a-b)$]

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

⇒$\frac{7 +4\sqrt{3} }{1} + \frac{7 -4 \sqrt{3} }{1} + \frac{4 - 2\sqrt{3} }{2}$

⇒$(7 + 4\sqrt{3} )+ (7 -4\sqrt{3} ) + (2 - \sqrt{3} )$

Now, we open the brackets,

⇒7 + $4\sqrt{3}$ + 7 - $4\sqrt{3}$ + 2 - $\sqrt{3}$

⇒16 - $\sqrt{3}$

Final answer:

Hence, the $16 - \sqrt{3}$ is the required value of $\frac{2 + \sqrt{3} }{2 - \sqrt{3} } + \frac{2 - \sqrt{3}}{2 + \sqrt{3} } + \frac{\sqrt{3}-1 }{\sqrt{3}+1 }$.

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