\(\begin{array}{l} a)\left( {x + 1} \right) + \left( {x + 3} \right) + \left( {x + 5} \right) + ... + \left( {x + 99} \right) = 0\\ \Leftrightarrow 50x + \left( {1 + 3 + 5 + ... + 99} \right) = 0\\ \Leftrightarrow 50x + \left( {99 + 1} \right).25 = 0\\ \Leftrightarrow 50x + 2500 = 0\\ \Leftrightarrow x = - 50 \end{array}\)
\(\begin{array}{l} b)\left( {x - 3} \right) + \left( {x - 2} \right) + \left( {x - 1} \right) + ... + 10 + 11 = 11\\ \Leftrightarrow \left( {x - 3} \right) + \left( {x - 2} \right) + \left( {x - 1} \right) + \left( {1 + 2 + 3 + ... + 10} \right) = 0\\ \Leftrightarrow \left( {x - 3} \right) + \left( {x - 2} \right) + \left( {x - 1} \right) + 55 = 0\\ \Leftrightarrow \left( {x - 3} \right) + \left( {x - 2} \right) + \left( {x - 1} \right) = - 55\\ \Leftrightarrow 3x = - 49\\ \Leftrightarrow x = - \dfrac{{49}}{3} \end{array}\)
a/ \(\left(x+1\right)+\left(x+3\right)+...+\left(x+5\right)+...+\left(x+99\right)=0\)
Số số hạng là: \(\frac{99-1}{2}=49\) (số hạng)
Ta có: \(\left(x+1\right)+\left(x+3\right)+...+\left(x+5\right)+...+\left(x+99\right)=0\)
=> \(50x+\left(1+3+...+99\right)=0\)
=> \(50x+\frac{\left(99+1\right).50}{2}=0\)
=> \(50x+2500=0\)
=> \(50x=0-2500=-2500\)
=> \(x=-2500:50=-50\)
Vậy: \(x=50\)
\(\begin{array}{l} c)x + \left( {x + 1} \right) + \left( {x + 2} \right) + ... + 2018 + 2019 = 2019\\ \Leftrightarrow x + \left( {x + 1} \right) + \left( {x + 2} \right) + ... + 2018 = 0 \end{array}\)
Số các số hạng: \(\dfrac{{2018 - x}}{1} + 1 = 2019 - x\)
Trung bình cộng: \(\dfrac{{2018 + x}}{2}\)
\(\begin{array}{l} \Rightarrow \dfrac{{\left( {2018 + x} \right)}}{2}.\left( {2019 - x} \right) = 0\\ \Leftrightarrow \left[ \begin{array}{l} 2018 + x = 0\\ 2019 - x = 0 \end{array} \right. \Leftrightarrow \left[ \begin{array}{l} x = - 2018\\ x = 2019 \text{(loại vì số hạng số bằng 0)} \end{array} \right. \end{array}\)
