\(B=\frac{1}{3.5}+\frac{1}{5.7}+...+\frac{1}{9.11}+\frac{1}{11.13}\)
\(2B=2\left(\frac{1}{3.5}+\frac{1}{5.7}+...+\frac{1}{11.13}\right)\)
\(2B=\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{11.13}\)
\(2B=\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{11}-\frac{1}{13}\)
\(2B=\frac{1}{3}-\frac{1}{13}\)
\(2B=\frac{10}{39}\)
\(B=\frac{5}{39}\)
toán trẻ trâu chưa đọc đã bít cách giải rùi
Nguyên Phương Yên sai rùi tui tính máy tính thử lại hẳn hoi nhé
\(B=\frac{1}{3.5}+\frac{1}{5.7}+...+\frac{1}{9.11}+\frac{1}{11.13}\)
\(2B=2\left(\frac{1}{3.5}+\frac{1}{5.7}+...+\frac{1}{11.13}\right)\)
\(2B=\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{11.13}\)
\(2B=\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{11}-\frac{1}{13}\)
\(2B=\frac{1}{3}-\frac{1}{13}\)
\(2B=\frac{10}{39}\)