bạn theo công thức nay nè
\(\frac{n}{a\left(a+n\right)}=\frac{1}{a}-\frac{1}{a+n}\)là giải đc thôi
\(\frac{1}{2x3}+\frac{1}{3x4}+\frac{1}{4x5}+...+\frac{1}{9x10}\) \(=\) \(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{9}-\frac{1}{10}\)
\(=\frac{1}{2}-\frac{1}{10}\)
\(=\frac{2}{5}\)
1/2x3 + 1/3x4 +1/4x5 +...+ 1/9x10
=1/2 - 1/3 + 1/3 - 1/4 + 1/4 - 1/5 +...+ 1/9 - 1/10
=1/2 - 1/10
=5/10 - 1/10
=4/10
=2/5
