Question

1/(sqrt(36)+sqrt(27))+1/(sqrt(27)+sqrt(18))+1/(sqrt(18)+sqrt(9))
I need help, idk how to begin with this problem

Answer 1

\(\dfrac{1}{\sqrt{36}+\sqrt{27}}+ \dfrac{1}{\sqrt{27}+\sqrt{18}}+\dfrac{1}{\sqrt{18}+\sqrt{9}}\)

Is this your question? You're supposed to simplify this?

Answer 2

yes

Answer 3

So let's simplify some things first

what is \(\sqrt{36}\)
what is \(\sqrt{9}\)

Answer 4

ok so its 6, and 3

Answer 5

Good so we have

\(\dfrac{1}{6+\sqrt{27}}+ \dfrac{1}{\sqrt{27}+\sqrt{18}}+\dfrac{1}{\sqrt{18}+3}\)

How can we re-write \(\sqrt{27}\) and \(\sqrt{18}\)

remember that 27 = 9*3
and 18 = 9*2

\(\sqrt{a\times b} = \sqrt{a}\times \sqrt{b}\)

and \(\sqrt{x^2} = x\)

so how can we simplify \(\sqrt{27}\) and \(\sqrt{18}\)

Answer 6

as 3sqrt(3) and 3sqrt(2)?

Answer 7

Exactly! So now we have

\(\dfrac{1}{6+3\sqrt{3}}+ \dfrac{1}{3\sqrt{3}+3\sqrt{2}}+\dfrac{1}{3\sqrt{2}+3}\)

Answer 8

Now the only way to add the fractions is to have the denominator

And the only way to get the same denominator is by multiply both the numerator and denominator by a term so that way you can add them

so for example,

1/x + 1/y

you multiply 1/x by y/y
you multiply 1/y by x/x

and you would get

x/(xy) + y/(xy)

and now you can add them and get (x+y)/ (xy)

Answer 9

So
\(\dfrac{1}{6+3\sqrt{3}}+ \dfrac{1}{3\sqrt{3}+3\sqrt{2}}+\dfrac{1}{3\sqrt{2}+3}\)

Let's add two fractions first and then simplify and then add the third one later


\(\dfrac{1}{6+3\sqrt{3}}+ \dfrac{1}{3\sqrt{3}+3\sqrt{2}}\)


\(=\left(\dfrac{1}{6+3\sqrt{3}}\times\dfrac{3\sqrt{3}+3\sqrt{2}}{3\sqrt{3}+3\sqrt{2}}\right) + \left(\dfrac{1}{3\sqrt{3}+3\sqrt{2}}\times\dfrac{6+3\sqrt{3}}{6+3\sqrt{3}}\right)\)

Answer 10

Can you simplify it? What do you get after you multiply it and then add them

Answer 11

so you would have
(3sqrt(3)+3sqrt(2)+6+3sqrt(3))/27+18√2+18√3+9√6=(6sqrt(3)+6+3sqrt(2))/(27+18√2+18√3+9√6 )

Answer 12

yes, I think that looks good so far

then you want to give that mess a common denominator with the 1/(3√2 + 3) chunk

Answer 13

\(\large \frac{6 \sqrt3 + 2 \sqrt2+6}{9 \sqrt6 + 18\sqrt3 + 18 \sqrt2 + 27}\times \frac{3\sqrt2+3}{3\sqrt2 +3}~~~+~~~\frac{1}{3\sqrt2 +3}\times\frac{9 \sqrt6 + 18\sqrt3 + 18 \sqrt2 + 27}{9 \sqrt6 + 18\sqrt3 + 18 \sqrt2 + 27}\)

Answer 14

tbh it might make our lives easier if we factored out a 1/3 from everything

Answer 15

oh how could i do that :p

Answer 16

\(\color{#0cbb34}{\text{Originally Posted by}}\) Angle
\(\large \frac{6 \sqrt3 + 2 \sqrt2+6}{9 \sqrt6 + 18\sqrt3 + 18 \sqrt2 + 27}\times \frac{3\sqrt2+3}{3\sqrt2 +3}~~~+~~~\frac{1}{3\sqrt2 +3}\times\frac{9 \sqrt6 + 18\sqrt3 + 18 \sqrt2 + 27}{9 \sqrt6 + 18\sqrt3 + 18 \sqrt2 + 27}\)
\(\color{#0cbb34}{\text{End of Quote}}\)

If we factor out a 1/3 maybe we'll get something like this
\(\large \frac{1}{3}(\frac{6 \sqrt3 + 2 \sqrt2+6}{3 \sqrt6 + 6\sqrt3 + 6 \sqrt2 + 9}\times \frac{\sqrt2+1}{\sqrt2 +1}~~~+~~~\frac{1}{\sqrt2 +1}\times\frac{3 \sqrt6 + 6\sqrt3 + 6 \sqrt2 + 9}{3 \sqrt6 + 6\sqrt3 + 6 \sqrt2 + 9})\)

Answer 17

there's a good chance I made a typo or something though @AZ

Answer 18

yeah, definitely a typo

Answer 19

In the numerator you accidentally put 2sqrt(2) instead of 3sqrt(2)

Answer 20

\(\color{#0cbb34}{\text{Originally Posted by}}\) Angle
\(\large \frac{6 \sqrt3 + 3 \sqrt2+6}{9 \sqrt6 + 18\sqrt3 + 18 \sqrt2 + 27}\times \frac{3\sqrt2+3}{3\sqrt2 +3}~~~+~~~\frac{1}{3\sqrt2 +3}\times\frac{9 \sqrt6 + 18\sqrt3 + 18 \sqrt2 + 27}{9 \sqrt6 + 18\sqrt3 + 18 \sqrt2 + 27}\)
\(\color{#0cbb34}{\text{End of Quote}}\)

If we factor out a 1/3 maybe we'll get something like this
\(\large \frac{1}{3}(\frac{6 \sqrt3 + 3 \sqrt2+6}{3 \sqrt6 + 6\sqrt3 + 6 \sqrt2 + 9}\times \frac{\sqrt2+1}{\sqrt2 +1}~~~+~~~\frac{1}{\sqrt2 +1}\times\frac{3 \sqrt6 + 6\sqrt3 + 6 \sqrt2 + 9}{3 \sqrt6 + 6\sqrt3 + 6 \sqrt2 + 9})\)

Answer 21

that makes more sense because then you can factor out a 3 in both the numerator and denominator to simplify it a bit if you wanted it

or we can brute for this

Answer 22

If we factor out a 1/3 maybe we'll get something like this
\(\large \frac{1}{3}(\frac{2 \sqrt3 + \sqrt2+2}{\sqrt6 + 2\sqrt3 + 2 \sqrt2 + 3}\times \frac{\sqrt2+1}{\sqrt2 +1}~~~+~~~\frac{1}{\sqrt2 +1}\times\frac{\sqrt6 + 2\sqrt3 + 2 \sqrt2 + 3}{\sqrt6 + 2\sqrt3 + 2 \sqrt2 + 3})\)

Answer 23

\(\color{#0cbb34}{\text{Originally Posted by}}\) @AZ
Exactly! So now we have

\(\dfrac{1}{6+3\sqrt{3}}+ \dfrac{1}{3\sqrt{3}+3\sqrt{2}}+\dfrac{1}{3\sqrt{2}+3}\)
\(\color{#0cbb34}{\text{End of Quote}}\)

Honestly, on second thought, it would probably be so much easier if we just rationalized the fractions

\( \dfrac{1}{6+3\sqrt{3}} = \dfrac{1}{6+3\sqrt{3}} \times \dfrac{6-3\sqrt{3}}{6-3\sqrt{3}} \)

\( \dfrac{6-3\sqrt{3}}{(6+3\sqrt{3})(6-3\sqrt{3})} = \dfrac{(6-3\sqrt{3})}{36-27} = \dfrac{6-3\sqrt{3}}{9} \)

and then similarly for the other two fractions

Answer 24

sorry, we're probably confusing you

Answer 25

wait wouldn't the one that AZ wrote be the same as before cus 6-3sqrt(3)/6sqrt(3) is the same as 1 or am i missing something

Answer 26

That is the same as 1

We're multiplying by that because it's 1 and we're trying to use the property (a-b) * (a+b) = a^2 - b^2

so that way we can get rid of the square roots in the denominator and bring it to the numerator

and you see that by doing that, we have a nice number in the denominator

Answer 27

Found a nice image that might explain it slightly better

Answer 28

(I assumed that they meant to write 6-3sqrt(3) but made a typo)

Answer 29

yea

Answer 30

I'm obviously not awake yet

Answer 31

wait so if i do the samething for 1/3sqrt(2)+3sqrt(3) i would get (3sqrt(2)-3sqrt(3))/-9 I think

Answer 32

yes

Answer 33

yeah that could be written as

(3sqrt(3) - 3sqrt(2)) / 9

Answer 34

if you wanted to get rid of that negative sign

Answer 35

so then for 1/(3sqrt(3)+3) it is (3sqrt(2)-3)/9

Answer 36

yes

Answer 37

and then i add them all together. cool. :)

Answer 38

so much easier than the common denominator mess ;-;

Answer 39

i got 3 and that over 9 is 1/3 thanks so much

Answer 40

now put it all together

\(\dfrac{6-3\sqrt{3}}{9} + \dfrac{3\sqrt{3} - 3\sqrt{2}}{9} + \dfrac{3\sqrt{2}-3}{9}\)

Answer 41

\(\color{#0cbb34}{\text{Originally Posted by}}\) @Angle
so much easier than the common denominator mess ;-;
\(\color{#0cbb34}{\text{End of Quote}}\)

yes ;-;
idk what I was thinking at first ;--;

Answer 42

\(\color{#0cbb34}{\text{Originally Posted by}}\) @crispyrat
i got 3 and that over 9 is 1/3 thanks so much
\(\color{#0cbb34}{\text{End of Quote}}\)

Good job!