Question

Tính giá trị của biểu thức

\(A=\frac{1}{2.5}+\frac{1}{5.8}+......+\frac{1}{32.35}\)

  Ai nhanh mk tick

Answer 1

\(A=\frac{1}{2\cdot5}+\frac{1}{5\cdot8}+...+\frac{1}{32\cdot35}\)

\(3A=\frac{3}{2\cdot5}+\frac{3}{5\cdot8}+...+\frac{3}{32\cdot35}\)

\(3A=\frac{1}{2}-\frac{1}{5}+\frac{1}{5}-\frac{1}{8}+...+\frac{1}{32}-\frac{1}{35}\)

\(3A=\frac{1}{2}-\frac{1}{35}\)

\(3A=\frac{33}{70}\)

\(A=\frac{11}{70}\)

Hok tốt !

Answer 2

Ta có : 

3A = \(\frac{3}{2.5}+\frac{3}{5.8}+....+\frac{3}{32.35}\)

=> 3A = \(\frac{1}{2}-\frac{1}{5}+\frac{1}{5}-\frac{1}{8}\)\(+\frac{1}{8}-....-\frac{1}{32}+\frac{1}{32}-\frac{1}{35}\)

=> 3A = \(\frac{1}{2}-\frac{1}{35}=\frac{33}{70}\)

=> A = \(\frac{11}{70}\)