Ngan Nguyen
Câu hỏi của Đặng Quý Dương - Toán lớp 6 - Học toán với OnlineMath
\(B=\dfrac{5}{2.1}+\dfrac{4}{1.11}+\dfrac{3}{11.2}+\dfrac{1}{2.15}+\dfrac{13}{15.4}\)
\(7B=\dfrac{7}{7}\left(\dfrac{5}{2.1}+\dfrac{4}{1.11}+\dfrac{3}{11.2}+\dfrac{1}{2.15}+\dfrac{13}{15.4}\right)\)
\(7B=\dfrac{5}{2.7}+\dfrac{4}{7.11}+\dfrac{3}{11.14}+\dfrac{1}{14.15}+\dfrac{13}{15.28}\)
\(7B=\dfrac{1}{2}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{11}+\dfrac{1}{11}-\dfrac{1}{14}+\dfrac{1}{14}-\dfrac{1}{15}+\dfrac{1}{15}-\dfrac{1}{28}\)
\(7B=\dfrac{1}{2}-\dfrac{1}{28}\)
\(7B=\dfrac{13}{28}\)
\(B=\dfrac{13}{4}\)
\(\dfrac{5.7}{2.7}\)+\(\dfrac{4.7}{7.11}\)+\(\dfrac{3.7}{11.14}\)+\(\dfrac{1.7}{14.15}\)+\(\dfrac{13.7}{15.28}\)
7(\(\dfrac{5}{2.7}\)+\(\dfrac{4}{7.11}\)+\(\dfrac{3}{11.14}\)+\(\dfrac{1}{14.15}\)+\(\dfrac{13}{15.28}\))
=7. (\(\dfrac{1}{2}\)-\(\dfrac{1}{7}\)+\(\dfrac{1}{7}\)-\(\dfrac{1}{11}\)+\(\dfrac{1}{11}\)-\(\dfrac{1}{14}\)+\(\dfrac{1}{14}\)-\(\dfrac{1}{15}\)+\(\dfrac{1}{15}\)-\(\dfrac{1}{28}\))
=7.(\(\dfrac{1}{2}\)-\(\dfrac{1}{28}\))
=7.\(\dfrac{14-1}{28}\)=7.\(\dfrac{13}{28}\)=\(\dfrac{13}{4}\)