Question

Tìm x ∈ Z , biết: (x+1)+(x+3)+(x+5)+(x+7)+...........+(x+99) = 0

Answer 1

x=-50

Answer 2

( x + 1 ) + ( x + 3 ) + ( x + 5 ) + ( x + 7 ) + .... + ( x + 99 ) = 0

=> ( x + x + x + ... + x ) + ( 1 + 3 + 5 + 7 + ... + 99 ) = 0

=> 50.x + 2500 = 0

=> 50.x = - 2500

=> x = - 50

Answer 3

Ta có: \(\left(x+1\right)+\left(x+3\right)+\left(x+5\right)+\left(x+7\right)+...+\left(x+99\right)=0\)

\(\Leftrightarrow x+1+x+3+x+5+x+7+...+x+99=0\)

\(\Leftrightarrow50x+2500=0\)

\(\Leftrightarrow50x=-2500\)

\(\Leftrightarrow x=-50\)(thỏa mãn)

Vậy: x=-50

Answer 4

(x + 1) + (x + 3) + (x + 5) + (x + 7) + .... + (x + 99) = 0

(x + x + .... + x) + (1 + 3 + 5 + 7 + .... + 99) = 0

(99 - 1) : 2 + 1 = 50

(99 + 1) . 50 : 2 = 2500

(x . 50) + 2500 = 0

x . 50 = 0 - 2500

x . 50 = -2500

x = -2500 : 50

= -50