Đáp án:
`\frac{67}{4104}`
Giải thích các bước giải:
$\dfrac{2}{4.7}-\dfrac{3}{5.9}+\dfrac{2}{7.10}-\dfrac{3}{9.13}+......+\dfrac{2}{301.304}-\dfrac{3}{401.405}$
$=(\dfrac{2}{4.7}+\dfrac{2}{7.10}+....+\dfrac{2}{301.304})-(\dfrac{3}{5.9}+\dfrac{3}{9.13}+....+\dfrac{3}{401.405})$
`=\frac{2}{3}.(\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+....+\frac{1}{301}-\frac{1}{304})`
`-\frac{3}{4}.(\frac{1}{5}-\frac{1}{9}+\frac{1}{9}-\frac{1}{13}+...+\frac{1}{401}-\frac{1}{405})`
`=\frac{2}{3}.(\frac{1}{4}-\frac{1}{304})-\frac{3}{4}.(\frac{1}{5}-\frac{1}{405})`
`=\frac{2}{3}.\frac{75}{304}-\frac{3}{4}.\frac{16}{81}`
`=\frac{25}{152}-\frac{4}{27}`
`=\frac{67}{4104}`
Chúc bạn học tốt !!!
