Question

\(\frac{12}{3x7}+\frac{12}{7x11}+\frac{12}{11x15}+..........+\frac{12}{39x43}+\frac{12}{43x47}\)

Các bn làm cả cách giải giúp mh nha !!!!!!!!!!!!!!!!!!!!!!!!

Answer 1

lớp 4 mà học dạng này rồi sao!?

Answer 2

\(\frac{12}{3\cdot7}+\frac{12}{7\cdot11}+...+\frac{12}{43\cdot47}=3\left(\frac{4}{3\cdot7}+\frac{4}{7\cdot11}+...+\frac{4}{43\cdot47}\right)=3\left(\frac{1}{3}-\frac{1}{7}+\frac{1}{7}-\frac{1}{11}+...+\frac{1}{43}-\frac{1}{47}\right)\)\(=3\left[\left(\frac{1}{3}-\frac{1}{47}\right)+\left(\frac{1}{7}-\frac{1}{7}\right)+...+\left(\frac{1}{43}-\frac{1}{43}\right)\right]=3\left[\left(\frac{47}{141}-\frac{3}{141}\right)+0+...+0\right]=3\cdot\frac{44}{141}=\frac{44}{47}\)

Answer 3

=\(3\cdot\left(\frac{4}{3\cdot7}+\frac{4}{7\cdot11}+...+\frac{12}{43\cdot47}\right)=3\left(\frac{1}{3}\left(\frac{1}{7}-\frac{1}{7}\right)-\left(\frac{1}{11}-\frac{1}{11}\right)-...-\frac{1}{47}\right)\)

=\(3\left(\frac{1}{3}-\frac{1}{47}\right)=3\cdot\frac{44}{141}=\frac{44}{47}\)

Nếu đúng cho xin lik nhé