Question

please answer this as soon as possible​

Answer 1

||✪✪ QUESTION ✪✪||

if (1/x+1) + (2/y+2) + (2006/z+2006) = 1, find the value of (x²/x²+x) + (y²/y²+2y) + (z²/z²+2006z)

|| ✰✰ ANSWER ✰✰ ||

(1/x+1) + (2/y+2) + (2006/z+2006) = 1

Adding and Subtracting 1 From each term in LHS, we get,

→ [ {(1/x+1) -1 } + 1 ] + [ {(2/y+2) -1 } + 1 ] + [ {(2006/z+2006) -1 } + 1 ] = 1

Taking LCM with (-1) now, we get,

→ [ {(1 - x - 1)/(x+1)} + 1 ] + [ {(2 - y - 2)/(y+2)} + 1 ] + [ {(2006 - z - 2006)/(z+2006)} + 1 ] = 1

→ {(-x) /(x+1)} + {(-y)/(y+2)} + {(-z)/(z+2006)} = 1 - 3

→ Taking (-ve) Common , or (-1) common From both sides now,

→ x/(x+1) + y(y+2) + z/(z+2006) = 2

Now Multiply and divide each Term by x , y and z in LHS, we get, ( or we can say we are Multiplying by 1).

→ (x/x){x/(x+1)} + (y/y){y(y+2)} + (z/z){z/(z+2006)} = 2

→ (x²/x²+x) + (y²/y²+2y) + (z²/z²+2006z) = 2 . (Ans).

Hence, Our Required Answer is 2.

Answer 2

Hence , (1) 2 is the answer.