cho M=17/5.-31/125.1/2.10/17.(-1/2)^3 và khi đó tổng M+N bằng

* Hướng dẫn giải

Đáp án cần chọn là: B

$\begin{array}{l}M=\frac{17}{5}.\frac{-31}{125}.\frac{1}{2}.\frac{10}{17}.{\left(\frac{-1}{2}\right)}^3\\ M=\frac{17.\left(-31\right)\mathrm{.1.10.}{\left(-1\right)}^3}{{\mathrm{5.125.2.17.2}}^3}\\ M=\frac{-31.\left(-1\right)}{{125.2}^3}\\ M=\frac{31}{1000}\end{array}$

$\begin{array}{l}N=\left(\frac{17}{28}+\frac{28}{29}-\frac{19}{30}-\frac{20}{31}\right).\left(\frac{-5}{12}+\frac{1}{4}+\frac{1}{6}\right)\\ N=\left(\frac{17}{28}+\frac{28}{29}-\frac{19}{30}-\frac{20}{31}\right).\left(\frac{-5}{12}+\frac{3}{12}+\frac{2}{12}\right)\\ N=\left(\frac{17}{28}+\frac{28}{29}-\frac{19}{30}-\frac{20}{31}\right).0\\ N=0\end{array}$

Vậy $M+N=\frac{31}{1000}+0=\frac{31}{1000}$