Sửa lại chút:
\(A=\frac{1}{1+2}+\frac{1}{1+2+3}+...+\frac{1}{1+2+3+..+10}=\frac{1}{3}+\frac{1}{6}+...+\frac{1}{55}=2x\left(\frac{1}{6}+\frac{1}{12}+\frac{1}{110}\right)\)
\(=2x\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{10}-\frac{1}{11}\right)=\frac{1}{2}x\left(\frac{1}{2}-\frac{1}{11}\right)=2x\frac{9}{22}=\frac{9}{11}\)