Bài 24:
` a) ` ` (x + 1) + (x + 3) + (x + 5) + ... + (x + 99) = 0 `
` => (x + x + x + ... + x) + (1 + 3 + 5 + ... + 99) = 0 `
` => 50x + 2500 = 0 `
` => 50x = -2500 `
` => x = -50 `
` b) ` ` (x - 3) + (x - 2) + (x - 1) + ... + 10 + 11 = 11 `
` <=> 3x + (3 - 3) + (2 - 2) + (1 - 1) + 4 + ... + 10 = 0 `
` <=> 3x + 4 + 5 + 6 + 7 + 8 + 9 + 10 = 0 `
` <=> 3x = 49 `
` <=> x = 49/3 `
` c) ` ` x + (x + 1) + (x + 2) + ... + 2018 + 2019 = 2019 `
` => x + (x + 1) + (x + 2) + ... + 2018 = 0 `
` => (2019 - x).\frac{2018+x}{2} = 0 `
` => ` \(\left[ \begin{array}{l}2019-x=0\\2018+x=0\end{array} \right.\)
` => ` \(\left[ \begin{array}{l}x=2019(loại)\\x=-2018\end{array} \right.\)
Vậy ` x = -2018 `