Question

tính tổng A=\(\frac{1}{51}+\frac{1}{52}+...+\frac{1}{99}+\frac{1}{100}\)

Answer 1

\(A=\frac{1}{51}+\frac{1}{52}+...+\frac{1}{99}+\frac{1}{100}\)

\(A=\frac{1}{51}+\frac{1}{52}+\frac{1}{51}+\frac{1}{99}+\frac{1}{100}\)

\(A=\left(\frac{1}{51}+\frac{1}{51}\right)+\frac{1}{52}+\frac{1}{99}+\frac{1}{100}\)

\(A=\frac{2}{51}+\frac{1}{52}+\frac{1}{99}+\frac{1}{100}\)

\(A=\frac{155}{1652}+\frac{1}{99}+\frac{1}{100}\)

\(A=\frac{5999}{87516}+\frac{1}{100}\)

\(A=0.078547465\)

Answer 2

mình mới lớp 6