Lời giải:
Ta có: F = \(\frac{0,125-\frac{1}{5}+\frac{1}{7}}{0,375-\frac{3}{5}+\frac{3}{7}}+\frac{\frac{1}{2}+\frac{1}{3}+0,2}{\frac{3}{4}+0,5+\frac{3}{10}}\) = \(\frac{0,125-\frac{1}{5}+\frac{1}{7}}{3.0,125-3.\frac{1}{5}+3.\frac{1}{7}}+\frac{\frac{1}{2}+\frac{1}{3}+\frac{1}{5}}{\frac{3}{2}.\frac{1}{2}+\frac{3}{2}.\frac{1}{3}+\frac{3}{2}.\frac{1}{5}}\)
= \(\frac{0,125-\frac{1}{5}+\frac{1}{7}}{3.\left(0,125-\frac{1}{5}+\frac{1}{7}\right)}+\frac{\frac{1}{2}+\frac{1}{3}+\frac{1}{5}}{\frac{3}{2}.\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{5}\right)}\)
= \(\frac{1}{3}+\frac{1}{\frac{3}{2}}\)
= \(\frac{1}{3}+\frac{2}{3}\)
= \(\frac{3}{3}\) = 1. Vậy: F = 1
Chúc bạn học tốt!
Tick cho mình nhé!
