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Question Number 40376 by behi83417@gmail.com last updated on 21/Jul/18

Commented by MrW3 last updated on 21/Jul/18
$(1) let u=x−2 (1/(u+1))+(3/(u−1))=1−(2/u) ((4u+2)/(u^2 −1))=((u−2)/u) 4u^2 +2u=u^3 −u−2u^2 +2 u^3 −6u^2 −3u+2=0 u=x−2=−0.8056, 0.3868, 6.4188 ⇒x=1.1944, 2.3868, 8.4188 (2) solution is: one of the three variables is a, the other two can be any value but with different + and − sign, e.g. { ((x=r=any real number)),((y=−r)),((z=a)) :}$
$(1)$ $l e t u = x - 2$ $\frac{1}{u + 1} + \frac{3}{u - 1} = 1 - \frac{2}{u}$ $\frac{4 u + 2}{u^{2} - 1} = \frac{u - 2}{u}$ $4 u^{2} + 2 u = u^{3} - u - 2 u^{2} + 2$ $u^{3} - 6 u^{2} - 3 u + 2 = 0$ $u = x - 2 = - 0.8056, 0.3868, 6.4188$ $\Rightarrow x = 1.1944, 2.3868, 8.4188$ $(2)$ $s o l u t i o n i s :$ $o n e o f t h e t h r e e v a r i a b l e s i s a, t h e$ $o t h e r t w o c a n b e a n y v a l u e b u t w i t h$ $d i f f e r e n t + a n d - s i g n, e . g .$ ${\begin{cases} x = r = a n y r e a l n u m b e r \\ y = - r \\ z = a \end{cases}$
Commented by behi83417@gmail.com last updated on 21/Jul/18

$\frac{1}{x} + \frac{1}{y} + \frac{1}{z} = \frac{1}{x + y + z} \Rightarrow$ $(x + y + z) (x y + y z + z x) - x y z = 0 \Rightarrow$ $(x + y) (y + z) (z + x) = 0 \Rightarrow$ $x = - y = r \in R, z = a .$
Commented by behi83417@gmail.com last updated on 21/Jul/18

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