\(B=\frac{5}{3}.\left(\frac{3}{1.4}+\frac{3}{4.7}+...+\frac{3}{100.103}\right)\)
\(B=\frac{5}{3}.\left(1-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{100}-\frac{1}{103}\right)\)
\(B=\frac{5}{3}.\left(1-\frac{1}{103}\right)\)
\(B=\frac{5}{3}.\frac{102}{103}=\frac{170}{103}\)
\(B=\frac{5}{1.4}+\frac{5}{4.7}+...+\frac{5}{100.103}\)
\(3B=5\left(\frac{3}{1.4}+\frac{3}{4.7}+...+\frac{3}{100.103}\right)\)
\(3B-B=5\left(1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+...+\frac{1}{100}-\frac{1}{103}\right)\)
\(2B=5\left(1-\frac{1}{103}\right)\)
\(2B=5.\frac{102}{103}\)
\(B=\frac{5.102}{103.2}\)
\(B=\frac{255}{103}\)
KL:...........................
xin lỗi b nhé. mk nhầm
b tham khảo bài của b trần tuấn anh nhé.
chúc bạn học tốt.
