bài này nâng cao lớp 6 mk giải rồi bạn nhờ ai giảng hộ nha nếu bn 5 lên 6
B=1/4.(4/1.5+4/5.9+......+4/25.29)
B=1/4.(1-1/5+1/5-1/9+.....+1/25-1/29)
B=1/4.(1-1/29)
B=1/4.28/29
B=7/29
\(B=\frac{1}{1.5}+\frac{1}{5.9}+\frac{1}{9.13}+\frac{1}{13.17}+\frac{1}{17.21}+\frac{1}{21.25}+\frac{1}{25.29}\)
\(\Rightarrow4B=\frac{4}{1.5}+\frac{4}{5.9}+\frac{4}{9.13}+\frac{4}{13.17}+\frac{4}{17.21}+\frac{4}{21.25}+\frac{4}{25.29}\)
\(\Rightarrow4B=1-\frac{1}{5}+\frac{1}{5}-\frac{1}{9}+\frac{1}{9}-\frac{1}{13}+\frac{1}{13}-\frac{1}{17}+\frac{1}{17}-\frac{1}{21}+\frac{1}{21}-\frac{1}{25}+\frac{1}{25}-\frac{1}{29}\)
\(\Rightarrow4B=1-\frac{1}{29}\)
\(\Rightarrow4B=\frac{29}{29}-\frac{1}{29}=\frac{28}{29}\)
\(\Rightarrow B=\frac{28}{29}:4=\frac{28}{29}.\frac{1}{4}=\frac{7}{29}\)
Vậy ....
\(B=\frac{1}{1\cdot5}+\frac{1}{5\cdot9}+\frac{1}{9\cdot13}+\frac{1}{13\cdot17}+\frac{1}{17\cdot21}+\frac{1}{21\cdot25}+\frac{1}{25\cdot29}\)
\(B=1-\frac{1}{5}+\frac{1}{5}-\frac{1}{9}+\frac{1}{9}-\frac{1}{13}+...+\frac{1}{25}-\frac{1}{29}\)
\(B=1-\frac{1}{29}\)
\(B=\frac{28}{29}\)
Chúc bạn học tốt !
B=1/1.5+1/5.9+1/9.13+....+1/25.29
B=1/4.(4/1.5 + 4/5.9 + 4/9.13 +... +1/25.29)
B=1/4.(1-1/5+1/5-1/9+1/9-1/13+...+1/25-1/29)
B=1/4.(1-1/29)
B=1/4.(29/29-1/29)
B=1/4.28/29
B=7/29
Vậy B=7/29
\(B=\frac{1}{1.5}+\frac{1}{5.9}+\frac{1}{9.13}+\frac{1}{13.17}+\frac{1}{17.21}+\frac{1}{21.25}+\frac{1}{25.29}\)
\(=\frac{1}{4}.\left(1-\frac{1}{5}+\frac{1}{5}-\frac{1}{9}+\frac{1}{9}-\frac{1}{13}+\frac{1}{13}-\frac{1}{17}+...+\frac{1}{25}-\frac{1}{29}\right)\)
\(=\frac{1}{4}.\left(1-\frac{1}{29}\right)\)
\(=\frac{1}{4}.\frac{28}{29}\)
\(=\frac{7}{29}\)
Vậy \(B=\frac{7}{29}\)
# Học tốt
\(B=\frac{1}{5.9}+\frac{1}{9.13}+\frac{1}{13.17}+\frac{1}{17.21}+\frac{1}{21.25}+\frac{1}{25.29}\)
\(=\frac{1}{4}\left(1-\frac{1}{5}+\frac{1}{5}-\frac{1}{9}+\frac{1}{9}-\frac{1}{13}+\frac{1}{13}-\frac{1}{17}+\frac{1}{17}-\frac{1}{21}+\frac{1}{21}-\frac{1}{25}+\frac{1}{25}-\frac{1}{29}\right)\)
\(=\frac{1}{4}\left(1-\frac{1}{29}\right)\)
\(=\frac{1}{4}.\frac{28}{29}\)
\(=\frac{7}{29}\)
