Mình sửa đề 1 chút
\(\frac{3}{1\cdot4}+\frac{3}{4\cdot7}+\frac{3}{7\cdot10}+....+\frac{3}{19\cdot22}\)
\(=1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+....+\frac{1}{19}-\frac{1}{22}\)
\(=1-\frac{1}{22}=\frac{20}{22}=\frac{10}{11}\)
3/1x4+3/4x7+3/7x10+...+3/19x22
=3/1-3/4+3/4-3/7+3/7-3/10+...+3/19-3/22
=1-1/4+1/4-1/7+1/7-1/10+...+1/19-1/22
=(1/4-1/4)+(1/7-1/7)+(1/10-1/10)+...+(1/19-1/19)+(1-1/22)
=0+0+0+...+0+21/22
=21/22
