\(S=20.21+22.23+23.24+....+39.40\)
\(3S=3.20.21+3.22.23+3.23.24+...+3.39.40\)
\(3S=20.21.\left(22-19\right)+22.23.\left(24-21\right)+....+39.40.\left(41-38\right)\)
\(3S=39.40.41-19.20.21\)
\(3S=55980\)
\(S=18660\).

Ta có: \(S=20.21+21.22+22.23+....+39.40\)

\(\Rightarrow3S=3.20.21+3.21.22+.....+3.39.40\)

\(\Rightarrow3S=20.21.\left(22-19\right)+22.23.\left(24-21\right)+.....+39.40.\left(41-38\right)\)

\(\Rightarrow3S=39.40.41-19.20.21\)

\(\Rightarrow3S=55980\)

\(\Rightarrow S=18660\)

Cách giải

=(39.40-20.21):0.01+1

=19.19:0.01+1

=1919+1

=1920

3s=20.21.(22-19)+21.22.(23-20)+22.23.(24-21)+...+39.40.(41-38)

=20.21.22 - 20.21.19+21.22.23 - 21.22.20 +22.23.24 -22.23.21 + ...+ 39.40.41-39.40.38

=39.40.41- 20.21.19=63960-7980=55980

s=18660

Ta có:

\(\frac{1}{20.21}+\frac{1}{21.22}+\frac{1}{22.23}+...+\frac{1}{60.61}\)

\(=\frac{1}{20}-\frac{1}{21}+\frac{1}{21}-\frac{1}{22}+\frac{1}{22}-\frac{1}{23}+...+\frac{1}{60}-\frac{1}{61}\)

\(=\frac{1}{2}-\frac{1}{61}=\frac{59}{122}\)

b) \(\frac{4}{5.9}+\frac{4}{9.13}+\frac{4}{13.17}+...+\frac{4}{45.49}\)

\(=\frac{1}{5.9}+\frac{1}{9.13}+\frac{1}{13.17}+...+\frac{1}{45.49}\)

\(=\frac{1}{5}-\frac{1}{9}+\frac{1}{9}-\frac{1}{13}+\frac{1}{13}-\frac{1}{17}+...+\frac{1}{45}-\frac{1}{49}\)

\(=\frac{1}{5}-\frac{1}{49}=\frac{44}{245}\)

Bn Tấn sai rùi

phần a , câu cuối là \(\frac{1}{20}\)chứ đâu phải \(\frac{1}{2}\)

Ta có: \(\frac{1}{20.21}=\frac{21-20}{20.21}=\frac{21}{20.21}-\frac{20}{20.21}=\frac{1}{20}-\frac{1}{21}\)

Suy ra: \(\frac{1}{20.21}+\frac{1}{21.22}+\frac{1}{22.23}+...+\frac{1}{60.61}\)

\(=\frac{1}{20}-\frac{1}{21}+\frac{1}{21}-\frac{1}{22}+\frac{1}{22}-\frac{1}{23}+...+\frac{1}{60}-\frac{1}{61}\)

\(=\frac{1}{20}-\frac{1}{61}=\frac{41}{1220}\)

Phép tính sau làm tương tự.