How to rationalize 1/1+ sqrt2 + sqrt3?

Question

answer: - sqrt6/4 + sqrt2/4 + 1/2
How do I get to there? I have never rationalized with 3 numbers before and I tried doing it by multiplying both sides to get rid of the square root but I just end up with a new problem all over again! I feel so limited just to one formula.. problem is that hteres the xtra variable. I duno how to get rid of the one, I tried doing like 1/1 + 1/sqrt2 + 1/sqrt3 and try to cancel sqrt invididually but that didnt seem to work either... iduno.. maybe I am just dumb at this

1/(1+sqrt2 +sqrt3) sorry

Answer 1

IS
1/1+ sqrt2 + sqrt3 = 1/(1+ sqrt2 + sqrt3) OR what


1/[(1+ sqrt2) + sqrt3] * [(1+ sqrt2) - sqrt3] / [(1+ sqrt2) - sqrt3]

= [(1+ √2) - √3] / [ -2-2√2]
= [(1+ √2) - √3] / [ -2-2√2]* [ -2+2√2] / [ -2+2√2]
= [(1+ √2) - √3]* [ -2+2√2]/ (-4)

X out the top and tidy up

Answer 2

it's a two step problem
think of 1/(1+sqrt2 +sqrt3) as
1/[(1+sqrt2) + sqrt3] and multiply by [(1+sqrt2) - sqrt3]/[(1+sqrt2) - sqrt3] to get rid of the sqrt3
taking advantage of the difference of perfect squares rule -
(a + b)(a - b) = (a^2 - b^2)
1/[(1+sqrt2) + sqrt3]*[(1+sqrt2) - sqrt3]/[(1+sqrt2) - sqrt3] =
[(1+sqrt2) - sqrt3]/[(1+sqrt2)^2 - (sqrt3)^2] =
[(1+sqrt2) - sqrt3]/[1+2sqrt2+ 2 - 3] =
[(1+sqrt2) - sqrt3]/2sqrt2 now multiply by (sqrt2/sqrt2)
[(1+sqrt2) - sqrt3]/2sqrt2* sqrt2/sqrt2 =
sqrt2(1 + sqrt2 - sqrt3)/2(sqrt2)^2 =
(sqrt2 + (sqrt2)^2 - sqrt2sqrt3)/2(2) =
(sqrt2 + 2 - sqrt6)/4 = sqrt2/4 + 2/4 - sqrt6/4 =
-sqrt6/4 + sqrt2/4 + 1/2

Answer 3

1/(1+ √2 + √3) =
(the trick is in the next step!)
= (1+√2-√3)/[(1+√2+√3) * (1+√2-√3)]
= (1+√2-√3) / (1+√2+√3+√2+2+√6-√3-√6–3)
= (1+√2-√3)/(2*√2)
= [(1+√2-√3)*√2]/[(2*√2)*√2]
= (√2+2-√6)/4
= -√6/4+ √2/4+½
-

Answer 4

=a million/(a million+sqrt 2 - sqrt 3) x [ (a million+ sqrt 2) + sqrt 3/ (a million+sqrt 2) + sqrt 3)] =a million + sqrt 2 + sqrt 3 / [ (a million+ sqrt 2)^2 -3 = a million + sqrt 2 + sqrt 3 / a million + 2 sqrt 2 + 2 - 3 = [a million + sqrt 2 + sqrt 3 / 2 sqrt 2] X sqrt 2/ sqrt 2 = sqrt 2 + 2 + sqrt 6 / 4

Answer 5

you have to rationalize twice

1/1 + sqrt(2) + sqrt(3) =

multiply with[1- (sqrt(2) + sqrt(3))]/[1 - (sqrt(2)+sqrt(3))]

1([1-(sqrt(2) + sqrt(3))]/ [(1 + sqrt(2) + sqrt(3))][(1-sqrt(2) - sqrt(3))]


([1-sqrt(2) - sqrt(3)]/ 1 -[(sqrt(2) + sqrt(3)]^2

([1-sqrt(2) - sqrt(3)]/ 1 - [2 + 3 + 2sqrt(6)]

([1-sqrt(2) - sqrt(3)]/ 1-[5 + 2sqrt(6)]

([1-sqrt(2) - sqrt(3))]/ - 4 - 2sqrt(6)

(-1/2)[1-sqrt(2) - sqrt(3)]/ (2+sqrt(6) )

now multiply with(2- sqrt(6))/(2 - sqrt(6))

-(1/2)[1-sqrt(2) - sqrt(3))(2 - sqrt(6))]/ (2 - sqrt(6) (2 +sqrt(6)

-(1/2)[2 - 2sqrt(2)-2sqrt(3)-sqrt(6)+sqrt(12)+sqrt(18))/(4-6)

-1/2[2 -2sqrt(2)-2sqrt(3)-sqrt(6)+2sqrt(3)+3sqrt(2) /(-2)

1/4[2 +sqrt(2) - sqrt(6)]