\(\frac{4}{3.6}+\frac{4}{6.9}+\frac{4}{9.12}+\frac{4}{12.15}\)
\(=\frac{4}{3}\left(\frac{3}{3.6}+\frac{3}{6.9}+\frac{3}{9.12}+\frac{3}{12.15}\right)\)
\(=\frac{4}{3}\left(\frac{1}{3}-\frac{1}{6}+\frac{1}{6}-\frac{1}{9}+\frac{1}{9}-\frac{1}{12}+\frac{1}{12}-\frac{1}{15}\right)\)
\(=\frac{4}{3}\left(\frac{1}{3}-\frac{1}{15}\right)\)
\(=\frac{4}{3}.\frac{4}{15}=\frac{16}{45}\)
giải:
\(\frac{4}{3x6}\)+\(\frac{4}{6x9}\)+\(\frac{4}{9x12}\)+ \(\frac{4}{12x15}\)
= \(\frac{4}{3}\)x(\(\frac{3}{3x6}\)+ \(\frac{3}{6x9}\)+\(\frac{3}{9x12}\)+\(\frac{3}{12x15}\))
=\(\frac{4}{3}\)x(1-\(\frac{1}{15}\))
=\(\frac{4}{3}\)x\(\frac{14}{15}\)
=\(\frac{56}{45}\)