Question

The fraction 2(root2+root6)\3(root2+root root3) is equal to....... plz i will mark as brainliest

Answer 1

2(√2 + √6 ) / 3 ( √2+ √√ 3)

Rationalize denominator,,

=>2(√2 + √6 ) (√2 - √√3) ÷ 3 ( √2+ √√ 3) (√2- √√3)

=>2 ( 2 - √(2√3) + √12 - √(6√3) )÷ 3 ( 2 - √3)

=>2 [ 2 - √(2√3) + 2√2 - √(6√3) ] ÷ 3* (2-√3)

=>(-2/3) [√2√2-√2√(√3) +2√2- √6√(√3) ] ÷ (2-√3)

=> rationalize again denominator
(-2/3) [√2√2-√2√(√3) +2√2- √6√(√3) ] (2+√3) ÷ (2-√3)(2+√3)

=> (-2/3) [√2√2-√2√(√3) +2√2- √6√(√3) ](2+√3) ÷ (4-3)

=> (-2/3) [ 2-√2√(√3) +2√2- √6√(√3) ] ( 2+ √3) ÷ 1

Let do multiplication in next step
, i will solve it on paper

Answer 2

Concept

Rationalization can be thought of as the process used to remove radicals or  imaginary numbers from the denominator of  algebraic fractions.

The mathematical conjugate of any binomial means another exact binomial with the opposite sign between the two terms.

Given

We have given a fraction  $\frac{2(\sqrt{2}+\sqrt{6}) }{3(\sqrt{2}+\sqrt{3}) }$ .

Find

We are asked to determine the value of the given fraction.

Solution

We need to rationalize the numerator as well as denominator of the given fraction.

Conjugate of numerator will be $\sqrt{2} -\sqrt{6}$ .

Conjugate of denominator will be $\sqrt{2} -\sqrt{3}$ .

Now, we will multiply the conjugate of numerator and denominator to the given fraction , we get

$\frac{2(\sqrt{2}+\sqrt{6})(\sqrt{2} -\sqrt{6} ) }{3(\sqrt{2}+\sqrt{3})(\sqrt{2} -\sqrt{3} ) }$

Applying identity in both numerator and denominator which is

$(a+b)(a-b)=a^2-b^2$

On applying identity we get,

$\frac{2(2-6)}{3(2-3)} \\\\=\frac{2(-4)}{3(-1)} \\\\=\frac{8}{3}$

Therefore, the given fraction is equal to 8/3.

#SPJ2