Cho P bằng bảy phần hai mươi cộng mười một phần mười lăm trừ mười lăm

* Hướng dẫn giải

Đáp án A

$$\begin{gathered} P=\left(\frac{7}{20}+\frac{11}{15}-\frac{15}{12}\right):\left(\frac{11}{20}-\frac{2}{45}\right) \\ P=\left(\frac{21}{60}+\frac{44}{60}-\frac{75}{60}\right):\left(\frac{99}{180}-\frac{104}{180}\right) \\ P=\frac{-10}{60}:\frac{-5}{180}=\frac{-10}{60}.\frac{180}{-5}=6 \end{gathered}$$

$$\begin{gathered} Q=\frac{5-{\displaystyle \frac{5}{8}}+{\displaystyle \frac{5}{9}}-{\displaystyle \frac{5}{27}}}{8-{\displaystyle \frac{8}{3}}+{\displaystyle \frac{8}{9}}-{\displaystyle \frac{8}{27}}}:\frac{15-{\displaystyle \frac{15}{11}}+{\displaystyle \frac{15}{121}}}{16-{\displaystyle \frac{16}{11}}+{\displaystyle \frac{16}{121}}} \\ Q=\frac{5\left(1-{\displaystyle \frac{1}{3}}+{\displaystyle \frac{1}{9}}-{\displaystyle \frac{1}{27}}\right)}{8.\left(1-\frac{1}{3}+\frac{1}{9}-\frac{1}{27}\right)}:\frac{15\left(1-{\displaystyle \frac{1}{11}}+{\displaystyle \frac{1}{121}}\right)}{16.\left(1-\frac{1}{11}+\frac{1}{121}\right)} \\ Q=\frac{5}{8}:\frac{15}{16}=\frac{5}{8}.\frac{16}{15}=\frac{2}{3} \end{gathered}$$

Vì  $6>\frac{2}{3}$ nên P > Q