# how to simplify-: `[(2+sqrt3)/(sqrt2+sqrt2+sqrt3)] +[(2-sqrt3)/(sqrt2-sqrt2-sqrt3)]`

lemjay | Certified Educator

`(2+sqrt3)/(sqrt2+sqrt2+sqrt3) + (2-sqrt3)/(sqrt2-sqrt2-sqrt3) `

Simplify the denominators of the two fractions by combining like terms.

= `(2+sqrt3)/(2sqrt2+sqrt3) + (2-sqrt3)/(-sqrt3) = (2+sqrt3)/(2sqrt2+sqrt3) - (2-sqrt3)/sqrt3`

The LCD of the two fractions is `sqrt3(2sqrt2+sqrt3)` . The corresponding equivalent fractions above are:

= `[sqrt3(2+sqrt3)]/[sqrt3(2sqrt2 + sqrt3)] - [(2-sqrt3)(2sqrt2+sqrt3)]/[sqrt3(2sqrt2+sqrt3)]`

= `(2sqrt3+3)/[sqrt3(2sqrt2+sqrt3)] - (4sqrt2+2sqrt3 -2sqrt6 - 3)/[sqrt3(2sqrt2+sqrt3)]`

= `[(2sqrt3 + 3) - (4sqrt2 + 2sqrt3 - 2sqrt6 - 3)]/[sqrt3(2sqrt2+sqrt3)]`

Combine like terms at the numerator.

= `(6 - 4sqrt2 + 2sqrt6) / (sqrt3(2sqrt2 + sqrt3)) = (6 - 4sqrt2 + 2sqrt6) / (2sqrt6 +3)`

Then,rationalize.

`=(6 - 4sqrt2 + 2sqrt6) / (2sqrt6+3) * (2sqrt6 - 3)/(2sqrt6-3) = (12sqrt6 - 8sqrt12 + 24 - 18 + 12sqrt2 - 6sqrt6)/(24-9)`

`=(12sqrt2 + 6sqrt6 - 8sqrt12 + 6)/15`

Answer: `(12sqrt2 + 6sqrt6 - 8sqrt12 + 6)/15`

sciencesolve | Certified Educator

You need to bring the terms to a common denominator, multiplying the first terms by `sqrt2- sqrt2- sqrt3`  and the next terms by `sqrt2 + sqrt2 + sqrt3`  such that:

`((2 + sqrt3)(sqrt2 - sqrt2 - sqrt3)+ (2 - sqrt3)(sqrt2 + sqrt2 + sqrt3))/((sqrt2 - sqrt2 - sqrt3)(sqrt2 + sqrt2 + sqrt3))`

You need to evaluate the possible eliminations to numerator such that:

`((2 + sqrt3)(-sqrt3) + (2 - sqrt3)(2sqrt2 + sqrt3))/((sqrt2)^2 - (sqrt2 + sqrt3)^2)`

`(-2sqrt3 - 3+ 4sqrt2 + 2sqrt3 - 2sqrt6 - 3)/(2 - (2 + 2sqrt6 + 3))`

`(-6 + 4sqrt2 + 2sqrt3 - 2sqrt6)/(2sqrt6 - 3)`

You need to multiply by the conjugate of denominator such that:

`((-6 + 4sqrt2 + 2sqrt3 - 2sqrt6)(2sqrt6+ 3))/(24 - 9)`

`(-12sqrt6 - 18 + 16sqrt3 + 12sqrt2 + 12sqrt2 + 6sqrt3 - 24 - 6sqrt6)/15`

`(-18sqrt6 + 22sqrt3 + 24sqrt2 - 42)/15`

Hence, performing the possible simplifications yields `(-18sqrt6 + 22sqrt3 + 24sqrt2 - 42)/15.`

jeew-m | Certified Educator

Let x= √2+√3

Then we can write (√2+√2+√3)= √2+x

Also (√2-√2-√3) = √2-(√2+√3) = √2-x

So (√2+√2+√3) (√2-√2-√3)= (√2+x)(√2-x)

= 2-x^2

= 2-(√2+√3)^2

= 2-(2+2*√2*√3+3)

= -(3+2√6)

[(2+√3)/(√2+√2+√3)] +[(2-√3)/(√2-√2-√3)]

= (2+√3)/(2+x) + (2-√3)/(2-x)

= [(2+√3)(2-x)+(2-√3)(2+x)]/[(√2+x)(√2-x)]

=[4-2x+2√3-√3*x+4+2x-2√3-√3*x]/[-(3+2√6)]

= [8-2√3*x]/[-(3+2√6)]

=[8-2√3*(√2+√3)]/[-(3+2√6)]

= [8-2√6-6]/[-(3+2√6)]

= 2(1-√6)/[-(3+2√6)]

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

= -2(1-√6)(3-2√6)/[(3+2√6)(3-2√6)]

= -2(3-5√6+12)/(9-4*6)

= -2(15-5√6)/-15

= 2*5(3-√6)/15

= 2(3-√6)/3