its easy!

Algebra Level 3

1 x + 1 + 2 y + 2 + 2006 z + 2006 = 1 f i n d x 2 x 2 + x + y 2 y 2 + 2 y + z 2 z 2 + 2006 z \frac { 1 }{ x+1 } +\frac { 2 }{ y+2 } +\frac { 2006 }{ z+2006 } =1\\ find\\ \frac { { x }^{ 2 } }{ { x }^{ 2 }+x } +\frac { { y }^{ 2 } }{ { y }^{ 2 }+2y } +\frac { { z }^{ 2 } }{ { z }^{ 2 }+2006z }

Assume that ( x 2 + x ) ( y 2 + 2 y ) ( z 2 + 2006 z ) ≠ 0 (x^2+x)(y^2+2y)(z^2+2006z) \neq 0 .


The answer is 2.

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Aryan Saxena by aryan saxena , 20, India

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3 solutions

Sujoy Roy
Dec 16, 2014

If the middle term is 2 y + 2 \frac{2}{y+2} , then adding left sides of two equations we get 3 3 . So, the value of the given expression is 3 − 1 = 2 3-1=\boxed{2} .

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It should be mentioned that x , y , z x,y,z are non-zero values because there are many cases where the given condition is met but the required value is undefined. Take, for example, the case of x = 0 , y = ( − 4 ) , z = 0 x=0,y=(-4),z=0 .

Assuming that they are non-zero values, we can do the following:

x 2 x 2 + x + y 2 y 2 + 2 y + z 2 z 2 + 2006 z = x x + 1 + y y + 2 + z z + 2006 = x + 1 − 1 x + 1 + y + 2 − 2 y + 2 + z + 2006 − 2006 z + 2006 = 1 + 1 + 1 − ( 1 x + 1 + 2 y + 2 + 2006 z + 2006 ) = 3 − 1 = 2 \small\begin{aligned}\frac{x^2}{x^2+x}+\frac{y^2}{y^2+2y}+\frac{z^2}{z^2+2006z}&=\frac{x}{x+1}+\frac{y}{y+2}+\frac{z}{z+2006}\\&=\frac{x+1-1}{x+1}+\frac{y+2-2}{y+2}+\frac{z+2006-2006}{z+2006}\\&=1+1+1-\left(\frac{1}{x+1}+\frac{2}{y+2}+\frac{2006}{z+2006}\right)\\&=3-1=2\end{aligned}

We obviously have ( x + 1 ) , ( y + 2 ) , ( z + 2006 ) ≠ 0 (x+1),(y+2),(z+2006)\neq 0 from the given condition which justifies the transition from step 2 to step 3.

Prasun Biswas - 6 years ago
Utkarsh Dwivedi
Dec 16, 2014

Hey , Aryan , well , by the initial statement did you meant this - 1 x + 1 + 2 y + 2 + 2006 z + 2006 = 1 \frac {1}{x+1} + \frac {2}{y+2}+\frac {2006}{z+2006} =1 or the one which you wrote in the question , I suppose my one , because I got the correct solution by my wrong interpretation , so do check it once and if wrong , edit .

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@Utkarsh Dwivedi , fixed, thanks for letting us know :D

Aditya Raut - 6 years, 5 months ago

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No problem.

Utkarsh Dwivedi - 6 years, 5 months ago

LET THE ANSWER BE x. THEN ADDING BOTH THE EQUATIONS WE GET 1+1+1 = 1+x that is x=2

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