Question

Plz help:)

Answer 1

If \[x=\frac{\sqrt{3}+\sqrt{2}}{\sqrt{3}-\sqrt{2}} \]& \[y=1.\]The value of \[\frac{x-y}{x-3y}\] is ?

Answer 2

rationalize x

Answer 3

Its answer is \[\color{red}{\frac{\sqrt{6}+4}{5}.}\]

Answer 4

I m nt getting my answer actually.
I have rationalised it

Answer 5

\[\huge{x=\frac{5+\sqrt{6}}{1}}\]
\[\huge{x=5+\sqrt{6}}\]

Answer 6

I have tried my best.

Answer 7

anybody tell me is the answer wrong?

Answer 8

\[\huge{\frac{4+\sqrt{6}}{2+\sqrt{6}}}\]
\[\huge{\frac{(4+\sqrt{6}(2-\sqrt{6})}{4-6}}\]
\[\huge{\frac{8-4\sqrt{6}+2\sqrt{6}-6}{-2}}\]
\[\huge{\frac{2-2\sqrt{6}}{-2}}\]
\[\huge{\frac{2\sqrt{6}-2}{2}}\]
\[\huge{\sqrt{6}-1}\]

Answer 9

@mathslover u wrote x=? wrong:/

Answer 10

i directly simplified that ...

Answer 11

x-y and x-3y ...

Answer 12

how? i think u should get\[5+2\sqrt{6}.\]

Answer 13

what is the answer ?

Answer 14

Ihave given it above

Answer 15

oh !!! sorry u r right

Answer 16

i have*

Answer 17

\[\huge{\frac{(4+2\sqrt{6})(2-2\sqrt{6})}{-20}}\]
\[\huge{\frac{8-4\sqrt{6}-24}{-20}}\]
\[\huge{\frac{-16-4\sqrt{6}}{-20}}\]
\[\huge{\frac{4\sqrt{6}+16}{20}}\]
\[\huge{\frac{\sqrt{6}+4}{5}}\]

Answer 18

k! thanx I gt it:)