The first equation of the set

Algebra Level 2

Let x , y , z x,y,z be the real numbers such that x y z = 1 xyz=1 . Find the value of

F = 1 1 + x + x y + 1 1 + y + y z + 1 1 + z + z x . F=\dfrac { 1 }{ 1+x+xy } +\dfrac { 1 }{ 1+y+yz } +\dfrac { 1 }{ 1+z+zx } .


The answer is 1.

Linkin Duck by Linkin Duck , 33, Vietnam

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2 solutions

Chew-Seong Cheong
Mar 24, 2017

F = 1 1 + x + x y + 1 1 + y + y z + 1 1 + z + z x = 1 1 + x + x y + x x + x y + x y z + x y x y + x y z + x y z x = 1 1 + x + x y + x x + x y + 1 + x y x y + 1 + x = 1 + x + x y 1 + x + x y = 1 \begin{aligned} F & = \frac 1{1+x+xy} + \frac 1{1+y+yz} + \frac 1{1+z+zx} \\ & = \frac 1{1+x+xy} + \frac {\color{#3D99F6}x}{{\color{#3D99F6}x}+{\color{#3D99F6}x}y+{\color{#3D99F6}x}yz} + \frac {\color{#D61F06}xy}{{\color{#D61F06}xy}+{\color{#D61F06}xy}z+{\color{#D61F06}xy}zx} \\ & = \frac 1{1+x+xy} + \frac x{x+xy+1} + \frac {xy}{xy+1+x} \\ & = \frac {1+x+xy}{1+x+xy} \\ & = \boxed{1} \end{aligned}

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Linkin Duck
Mar 24, 2017

Since x y z = 1 xyz=1 , we have:

1 1 + y + y z = x x + x y + x y z = x x + x y + 1 , \frac { 1 }{ 1+y+yz } =\frac { x }{ x+xy+xyz } =\frac { x }{ x+xy+1 } ,

1 1 + z + z x = x y x y + x y z + x 2 y z = x y x y + 1 + x . \frac { 1 }{ 1+z+zx } =\frac { xy }{ xy+xyz+{ x }^{ 2 }yz } =\frac { xy }{ xy+1+x } .

Hence, F = 1 + x + x y 1 + x + x y = 1 . F=\frac { 1+x+xy }{ 1+x+xy } =\boxed { 1 } .

1 Helpful 0 Interesting 0 Brilliant 0 Confused

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