Bài 1: \(\frac{x+1}{10}+\frac{x+1}{11}+\frac{x+1}{12}=\frac{x+1}{13}+\frac{x+1}{14}\)
\(\frac{x+1}{10}+\frac{x+1}{11}+\frac{x+1}{12}-\frac{x+1}{13}-\frac{x+1}{14}=0\)
\(\left(x+1\right).\left(\frac{1}{10}+\frac{1}{11}+\frac{1}{12}-\frac{1}{13}-\frac{1}{14}\right)=0\)
\(\text{Ta thấy }\frac{1}{10}+\frac{1}{11}+\frac{1}{12}-\frac{1}{13}-\frac{1}{14}\ne0\text{ nên:}\)
\(x+1=0\)
\(x=-1\)
Bài 2: \(\frac{-1}{100.99}-\frac{1}{99.98}-\frac{1}{98.97}-...-\frac{1}{3.2}-\frac{1}{2.1}\)
\(=-\left(\frac{1}{99}-\frac{1}{100}\right)-\left(\frac{1}{98}-\frac{1}{99}\right)-\left(\frac{1}{97}-\frac{1}{98}\right)-...-\left(\frac{1}{2}-\frac{1}{3}\right)-\left(\frac{1}{1}-\frac{1}{2}\right)\)
\(=-\frac{1}{99}+\frac{1}{100}-\frac{1}{98}+\frac{1}{99}-\frac{1}{97}+\frac{1}{98}-...-\frac{1}{2}+\frac{1}{3}-\frac{1}{1}+\frac{1}{2}\)
\(=\frac{1}{100}-\frac{1}{1}=\frac{1}{100}-\frac{100}{100}=\frac{-99}{100}\)
