Question

plz help.....

Answer 1

If x=\[\frac{a-b}{a+b},y=\frac{b-c}{b+c} and z=\frac{c-a}{c+a}\]
then the value of
\[\frac{(1+x)(1+y)(1+z)}{(1-x)(1-y)(1-z)}\]

Answer 2

plzz help me

Answer 3

u know componendo -dividendo?

Answer 4

@hartnn @robtobey @ganeshie8 @mathslover

Answer 5

no @hartnn

Answer 6

@Outkast3r09 @AccessDenied @Vincent-Lyon.Fr @akash_809 @Australopithecus @Algebraic!

Answer 7

\(\frac{(a+b)}{(c+d)}\) some thing like that.

Answer 8

@countonme123 @curiousshubham @eSpeX @nphuongsun93 @nincompoop

Answer 9

no

Answer 10

\(\huge if \frac{p}{q}= \frac{r}{s} then \quad \frac{p+q}{p-q}= \frac{r+s}{r-s}\)

Answer 11

apply this to x= (a-b)/(a+b)

Answer 12

what u get ?

Answer 13

Oh yes that's the : Componendo and Dividendo

Answer 14

a/b

Answer 15

right

Answer 16

that is correct, but what is on left side ?

Answer 17

a-b/a+b

Answer 18

no, i meant in terms of x
u got a/b on right side.
on left side ?

Answer 19

x

Answer 20

nope, and a/b is also incorrect, it should be -a/b

Answer 21

ok

Answer 22

then

Answer 23

didn't u get (x+1)/(x-1) on left ?

Answer 24

why

Answer 25

howz that

Answer 26

Lets go for numerator at first:
\[(1+x) + (1+y) + (1+z)\]\[(1 +\frac{ a-b }{ a+b }) + (1+\frac{ b-c }{ b+c }) + (1+\frac{ c-a }{ c+a }\]\[(\frac{ a+b+a-b }{ a+b }) (\frac{ b+c +b-c }{ b+c }) (\frac{ c+a+c-a }{ c+a })\]i.e. \[(\frac{ 2a }{ a+b })(\frac{ 2b }{ b+c })(\frac{ 2c }{ c+a })\]

Answer 27

\(\huge x=\frac{a-b}{a+b}\implies \frac{x+1}{x-1}=\frac{a-b+a+b}{a-b-a-b}\)
got that ?

Answer 28

ok

Answer 29

so (1+x)/(1-x) will be ?

Answer 30

a/b or -a/b ?

Answer 31

no

Answer 32

i don't understand

Answer 33

which part ?

Answer 34

so (1+x)/(1-x) will be ?

Answer 35

Now for denominator:\[(1-x) (1-y)(1-z)\]\[(1-\frac{ a-b }{ a+b })(1-\frac{ b-c }{ b+c })(1-\frac{ c-a }{ c+a })\]\[(\frac{ a+b-a+b }{ a+b })(\frac{ b+c-b+c }{ b+c })(\frac{ c+a-c+a }{ c+a })\]i.e.\[(\frac{ 2b }{ a+b })(\frac{ 2c }{ b+c })(\frac{ 2a }{ c+a })\]

Answer 36

ok

Answer 37

u said u got this.
\(\huge x=\frac{a-b}{a+b}\implies \frac{x+1}{x-1}=\frac{a-b+a+b}{a-b-a-b}\)
so now couldn't u find
(1+x)/(1-x) from this ?

Answer 38

So the numerator and denominator are the same i.e. it will cancel out. The simplification result will be 1.

Answer 39

thnx.... @curiousshubham
and also great help also my tutor @hartnn very thnx

Answer 40

you are welcomed!