It will make you crazy!!!

Algebra Level 4

If 1 x + 1 + 2 y + 2 + 2006 z + 2006 = 1 \frac{1}{x+1}+\frac{2}{y+2}+\frac{2006}{z+2006}=1 Find the value of x 2 x 2 + x + y 2 y 2 + 2 y + z 2 z 2 + 2006 z \frac{x^{2}}{x^{2}+x}+\frac{y^{2}}{y^{2}+2y}+\frac{z^{2}}{z^{2}+2006z}


The answer is 2.

Rayyan Shahid by Rayyan Shahid , 20, India

This section requires Javascript.
You are seeing this because something didn't load right. We suggest you, (a) try refreshing the page, (b) enabling javascript if it is disabled on your browser and, finally, (c) loading the non-javascript version of this page . We're sorry about the hassle.

2 solutions

U Z
Oct 26, 2014

x 2 x 2 + x + y 2 y 2 + 2 y + z 2 z 2 + 2006 z \frac{x^{2}}{x^{2} + x} + \frac{y^{2}}{y^{2} + 2y} + \frac{z^{2}}{z^{2} + 2006z}

= x x + 1 + y y + 2 + z z + 2006 = \frac{x}{x + 1} + \frac{y}{ y +2} + \frac{z}{z + 2006}

= x + 1 − 1 x + 1 + y + 2 − 2 y + 2 + z + 2006 − 2006 z + 2006 = \frac{ x + 1 - 1}{x + 1} + \frac{y + 2 - 2}{ y + 2} + \frac{z + 2006 - 200 6}{z + 2006}

= 3 − ( 1 x + 1 + 2 y + 2 + 2006 z + 2006 ) = 3 - ( \frac{ 1}{x + 1} + \frac{ 2}{ y + 2} + \frac{2006}{ z +2006})

= 3 − 1 = 2 = 3 -1 = 2

12 Helpful 0 Interesting 0 Brilliant 0 Confused

we can also assume each term to be 1/3 which will give x=2 y=4 z=4012

Mehul Chaturvedi - 6 years, 7 months ago

Log in to reply

thts how AD sir taught us

Prashant Ramnani - 6 years, 7 months ago

Log in to reply

Yeah prashant (the worst method)

Mehul Chaturvedi - 6 years, 7 months ago
Irtaza Sheikh
Apr 30, 2015

We can do it by consideration and it ll be easy for everyone . I just done this question in 3 - 4 minutes by consideration.
If you see the first condition all are positive then one of them should be negative.... .

0 Helpful 0 Interesting 0 Brilliant 0 Confused

0 pending reports

×

Problem Loading...

Note Loading...

Set Loading...