Question

Cho đẳng thức 100A=\(\frac{99}{1}+\frac{98}{2}+\frac{97}{3}+...+\frac{1}{99}\). Chứng minh rằng A không phải là số tự nhiên

Answer 1

100A = \(\frac{99}{1}+1+\frac{98}{2}+1+...+\frac{1}{99}+1-99\)

100A=\(\frac{100}{1}+\frac{100}{2}+\frac{100}{3}+...+\frac{100}{99}-99\)

100A =\(\left(\frac{100}{2}+\frac{100}{3}+..+\frac{100}{99}+100-99\right)\)

100A=\(\left(\frac{100}{2}+\frac{100}{3}+...+\frac{100}{99}+1\right)\)

100A=\(\left(\frac{100}{2}+\frac{100}{3}+...+\frac{100}{99}+\frac{100}{100}\right)\)

100A=100.\(\left(\frac{1}{2}+\frac{1}{3}+...+\frac{1}{99}+\frac{1}{100}\right)\)

A=\(\frac{1}{2}+\frac{1}{3}+...+\frac{1}{99}+\frac{1}{100}\)