If 1-x-1-2-y-z- 2006-2006-1- find the value of x2-x2-x-y2-y2-y - z2-z2-2006z

The correct option is C 2 x^2/(x^2+x) + y^2/(y^2+2y) +z^2/(z^2+2006z) nbsp; = x/(x+1) + y/ (y+2) + z/(z+2006) = (x+1 1)/(x+1) + (y+2 2)/(y+2) + (z+2 ...

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Byju's Answer

Standard VIII

Mathematics

Algebraic Identities

If 1/x+1+2/...

Question

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

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Solution

The correct option is C 2
$\frac{x^{2}}{(x^{2} + x)} + \frac{y^{2}}{(y^{2} + 2 y)} + \frac{z^{2}}{(z^{2} + 2006 z)}$
$= \frac{x}{(x + 1)} + \frac{y}{(y + 2)} + \frac{z}{(z + 2006)} = \frac{(x + 1 - 1)}{(x + 1)} + \frac{(y + 2 - 2)}{(y + 2)} + \frac{(z + 2006 - 2006)}{(z + 2006)}$
$= 1 - \frac{1}{(x + 1)} + 1 - \frac{2}{(y + 2)} + 1 - \frac{2006}{(z + 2006)}$
$= 3 - (\frac{1}{(x + 1)} + \frac{2}{(y + 2)} + \frac{2006}{(z + 2006)})$
Given $\frac{1}{(x + 1)} + \frac{2}{(y + 2)} + \frac{2006}{(z + 2006)} = 1$
$= 3 - 1 = 2$

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If 1x+1+2y+z+20062006=1, find the value of x2x2+x+y2y2+y+z2z2+2006z

The correct option is C 2x2(x2+x)+y2(y2+2y)+z2(z2+2006z) =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(x+1)+1−2(y+2)+1−2006(z+2006) =3−(1(x+1)+2(y+2)+2006(z+2006)) Given 1(x+1)+2(y+2)+2006(z+2006)=1 =3−1=2

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