Question

rationalize the denominator and simplify 4+√6/4-√3+4-√6/4+√6​

Answer 1

This is the answer to ur question... hope it helps

Answer 2

$\frac{4+\sqrt6}{4-\sqrt3}+\frac{4-\sqrt6}{4+\sqrt6}=\frac{3020+40\sqrt3+30\sqrt{2}-1000\sqrt6}{130}$

Step-by-step explanation:

Given : Expression $\frac{4+\sqrt6}{4-\sqrt3}+\frac{4-\sqrt6}{4+\sqrt6}$

To find : Rationalize the denominator and simplify ?

Solution :

$\frac{4+\sqrt6}{4-\sqrt3}+\frac{4-\sqrt6}{4+\sqrt6}$

Rationalize the denominator,

$=\frac{4+\sqrt6}{4-\sqrt3}\times \frac{4+\sqrt3}{4+\sqrt3}+\frac{4-\sqrt6}{4+\sqrt6}\times \frac{4-\sqrt3}{4-\sqrt3}$

$=\frac{(4+\sqrt6)(4+\sqrt3)}{4^2-(\sqrt3)^2}+\frac{(4-\sqrt6)^2}{4^2-(\sqrt6)^2}$

$=\frac{16+4\sqrt3+4\sqrt6+\sqrt{18}}{16-3}+\frac{16+6-8\sqrt6}{16-6}$

$=\frac{16+4\sqrt3+4\sqrt6+3\sqrt{2}}{13}+\frac{22-8\sqrt6}{10}$

$=\frac{(16+4\sqrt3+4\sqrt6+3\sqrt{2})10+(22-8\sqrt6)13}{130}$

$=\frac{160+40\sqrt3+40\sqrt6+30\sqrt{2}+2860-1040\sqrt6}{130}$

$=\frac{3020+40\sqrt3+30\sqrt{2}-1000\sqrt6}{130}$

Therefore, $\frac{4+\sqrt6}{4-\sqrt3}+\frac{4-\sqrt6}{4+\sqrt6}=\frac{3020+40\sqrt3+30\sqrt{2}-1000\sqrt6}{130}$

#Learn more

Rationalise the denominator sqrt3-4sqrt2/sqrt3-sqrt2

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