Question

pls HELP..... answer the second part of the 25th question in the attached paper.....

Answer 1

you want a^2+b^2-5ab ?

Answer 2

ya

Answer 3

just to make is easy ! let's calculate ab first
\[\frac{\sqrt{3}-\sqrt{2}}{\sqrt{3}+\sqrt{2}}.\frac{\sqrt{3}+\sqrt{2}}{\sqrt{3}-\sqrt{2}}\]
can you do that .?

Answer 4

is it = 2

Answer 5

no ! why 2 .?

Answer 6

notice that \[a=\frac{1}{b}\]
do you see that ?

Answer 7

just answer please ! @basith

Answer 8

id nt get it ....:(

Answer 9

ya i notice that....

Answer 10

now if you see that \[a=\frac{1}{b}\]
ab= ...?

Answer 11

can u show me all of i step by step.....then ill understand it better

Answer 12

kk ! np at all ! just I advise you to do exercises by yourself ! then you'll learn how to solve them !

Answer 13

ill understand it better because im used to studying like this...........first ill understand the steps by lookin....then try othere quest.. like it

Answer 14

\[a=\frac{1}{b} ====>ab=1\text{"we multiplied by "b" both sides"}\]
then 5ab=5

Answer 15

now let's calculate a^2 :)

Answer 16

first we can "simplify" a !

Answer 17

equation will be more help instead of ...like a^2......sry 4 he trouble

Answer 18

hw is a a=1/b ??????

Answer 19

do you know that
(a+b)(a-b)= a^2-b^2

it is very useful for this problem
also, multiply these expressions using FOIL : just as if it were (x+y)*(x+y)

Answer 20

\[a=\frac{\sqrt{3}-\sqrt{2}}{\sqrt{3}+\sqrt{2}} ===>a=\frac{(\sqrt{3}-\sqrt{2})(\sqrt{3}-\sqrt{2})}{(\sqrt{3}+\sqrt{2})(\sqrt{3}-\sqrt{2})}\]
\[\text{using (x-y)(x+y)=(}x^2-y^2)\]
we rationalize the denominator
a become \[a=\frac{(\sqrt{3}-\sqrt{2})^2}{{\sqrt{3}^{2}-\sqrt{2}}^2"=3-2=1"}\]
then a= \[a=(\sqrt{3}-\sqrt{2})^2\]
can you calculate a ?

Answer 21

can you do that ?

Answer 22

\[(\sqrt{3}^{2}-2\sqrt{3}\sqrt{2}+\sqrt{2^{2}})\]

Answer 23

=

Answer 24

\[3-2\sqrt{3}\sqrt{2}+2\]

Answer 25

= \[5-2\sqrt{3}\sqrt{2}\]

Answer 26

am i correct???

Answer 27

good !
now you can easily calculate : a ^2 yeaah ?

Answer 28

\[(5-2\sqrt{3}\sqrt{2})^{2}\]

Answer 29

=\[25-24= 1\]

Answer 30

right??

Answer 31

(x-y)^2=x^2+y^2-2xy

Answer 32

ya..so.....\[(5)^{2}-2\times5\times2\sqrt{3}\sqrt{2}+(2\sqrt{3}\sqrt{2})^{2}\]

Answer 33

=\[25-20\sqrt{3}\sqrt{2}+24\]

Answer 34

=\[49-20\sqrt{3}\sqrt{2}\]

Answer 35

/???right

Answer 36

I have to go ! Can you do the same for b ?

Answer 37

ill try ...thnx 4 the help ^_^

Answer 38

hw to find 5ab

Answer 39

did you see that \[\frac{1}{\frac{\sqrt{3}-\sqrt{2}}{\sqrt{3}+\sqrt{2}}}\\ \text{can you calculate this?}\]

Answer 40

idid nt under
stand...

Answer 41

just can you calculate what I typed ?!

Answer 42

just answer to my question ! Don't think about the exercise :) :) !

Answer 43

if we do the calculations we shud get \[1\div 49-20\sqrt{3}\sqrt{2}\]

Answer 44

with no calculation please ! Just \[\frac{1}{\frac{x}{y}}=\frac{y}{x}\]

Answer 45

did you see it now ?

Answer 46

isnt 1

Answer 47

wait....1min

Answer 48

ya......i got it

Answer 49

then???

Answer 50

ab=?!

Answer 51

donno :(

Answer 52

?????

Answer 53

\[\frac{1}{b}=a\]
did you see that ?

Answer 54

hw is 1/b = a

Answer 55

ohh ! I told you that \[\frac{1}{\frac{x}{y}}=\frac{y}{x}\]
do you see that is like "x/y" !
if you diidn't see that ...I can show something else kk ?!

Answer 56

do you see that b is in form x/y

Answer 57

yes....but where does 1 cm from

Answer 58

kk forget about that ! I'll show you something else kk ?!

Answer 59

k

Answer 60

\[ab=\frac{(\sqrt{3}+\sqrt{2})}{(\sqrt{3}-\sqrt{2})}.\frac{(\sqrt{3}-\sqrt{2})}{(\sqrt{3}+\sqrt{2})}\]
can you calculate this ?

Answer 61

do you see that the nominator of the first one is the denominator of the second one ?

Answer 62

yes..

Answer 63

and also the nominator of the second is denominator of the first !
what can you do in this case ?!

Answer 64

cn u tell me.....

Answer 65

im nt good at this

Answer 66

I have to go now ! I'll call for help ! kk ?

Answer 67

k

Answer 68

thnx

Answer 69

kk ! I'll give an example
\[\frac{5}{4}.\frac{4}{5}\] =??

Answer 70

1

Answer 71

yes ! How did you do that ?!

Answer 72

It's the same for ab ....see now ?!

Answer 73

so thats 5ab = 5 vright????

Answer 74

yeeah !

Answer 75

wait a sec.....

Answer 76

so according to the equation after the calculations we get,
\[a ^{2}+b ^{2}-5ab\]
=\[49-20\sqrt{3}\sqrt{2}+49+20\sqrt{3}2\sqrt{2}-5\times1\]
=\[98-5\]
=93....am i correct???

Answer 77

I think so ! It seems correct ! Wait I'll check again !

Answer 78

k

Answer 79

yeah ! that 's correct !

Answer 80

couldnt hv done it without u........thnx a lot

Answer 81

you're welcome ! :) !