a, Rút gọn hai phân số, ta có:

\(\frac{-2}{10}=\frac{-1}{5};\frac{8}{-20}=\frac{-8}{20}=\frac{-2}{5}\)

Mà: \(\frac{-1}{5}>\frac{-2}{5}\)

\(\Rightarrow\frac{-2}{10}>\frac{8}{-20}\)

b) Rút gọn phân số thứ hai, ta có:

\(\frac{-40}{-28}=\frac{40}{28}=\frac{10}{7}\)

Mà: \(\frac{10}{7}=\frac{10}{7}\)

\(\Rightarrow\frac{10}{7}=\frac{-40}{-28}\)

Ta có:

$S=\dfrac{5}{1\cdot2\cdot3}+\dfrac{8}{2\cdot3\cdot4}+\dfrac{11}{3\cdot4\cdot5}+\cdots+\dfrac{6026}{2008\cdot2009\cdot2010}$

Nhận thấy tử số có dạng $3n+2$, nên:

$S=\sum_{n=1}^{2008}\dfrac{3n+2}{n(n+1)(n+2)}$

Ta có: $\dfrac{3n+2}{n(n+1)(n+2)}=\dfrac{1}{n(n+1)}+\dfrac{2}{(n+1)(n+2)}$

Do đó:

$S=\left(\dfrac1{1\cdot2}+\dfrac1{2\cdot3}+\cdots+\dfrac1{2008\cdot2009}\right)$

$+2\left(\dfrac1{2\cdot3}+\dfrac1{3\cdot4}+\cdots+\dfrac1{2009\cdot2010}\right)$

Mà: $\dfrac1{n(n+1)}=\dfrac1n-\dfrac1{n+1}$

Nên: $S=\left(1-\dfrac1{2009}\right)+2\left(\dfrac12-\dfrac1{2010}\right)$

$=1-\dfrac1{2009}+1-\dfrac1{1005}$

$=2-\dfrac1{2009}-\dfrac1{1005}$

Vì: $\dfrac1{2009}+\dfrac1{1005}>0$

Nên $S<2$

nhanh lên các bn m còn 30 p nữa sắp phải nộp rùi

ai trả lời nhanh nhất m k cho

a) \(\frac{3}{5}:\left(-\frac{1}{15}-\frac{1}{6}\right)+\frac{3}{5}:\left(-\frac{1}{3}-1\frac{1}{15}\right)\)

\(=\frac{3}{5}:\left(-\frac{1}{15}-\frac{1}{6}-\frac{2}{6}-1+\frac{1}{15}\right)\)

\(=\frac{3}{5}:\left(-\frac{1}{2}-1\right)\)

\(=\frac{3}{5}:\left(-\frac{3}{2}\right)\)

\(=-\frac{2}{5}\)

b) \(\left(-\frac{3}{4}+\frac{5}{13}\right):\frac{2}{7}-\left(2\frac{1}{4}+\frac{8}{13}\right):\frac{2}{7}\)

\(=\left(-\frac{3}{4}+\frac{5}{13}-2+\frac{1}{4}+\frac{8}{13}\right):\frac{2}{7}\)

\(=\left(-\frac{1}{2}+1-2\right):\frac{2}{7}\)

\(=\left(-\frac{1}{2}-1\right):\frac{2}{7}\)

\(=-\frac{3}{2}:\frac{2}{7}\)

\(=-\frac{21}{4}\)