Cho 

- Hướng dẫn giải

Đáp án cần chọn là: D

$\begin{array}{l}M=\left(\frac{1}{3}+\frac{12}{67}+\frac{13}{41}\right)-\left(\frac{79}{67}-\frac{28}{41}\right)\\ M=\frac{1}{3}+\frac{12}{67}+\frac{13}{41}-\frac{79}{67}+\frac{28}{41}\\ M=\frac{1}{3}+\left(\frac{12}{67}-\frac{79}{67}\right)+\left(\frac{13}{41}+\frac{28}{41}\right)\\ M=\frac{1}{3}+\left(-1\right)+1\\ M=\frac{1}{3}\end{array}$

$\begin{array}{l}N=\frac{38}{45}-\left(\frac{8}{45}-\frac{17}{51}-\frac{3}{11}\right)\\ N=\frac{38}{45}-\frac{8}{45}+\frac{17}{51}+\frac{3}{11}\\ N=\left(\frac{38}{45}-\frac{8}{45}\right)+\frac{17}{51}+\frac{3}{11}\\ N=\frac{2}{3}+\frac{1}{3}+\frac{3}{11}\\ N=1+\frac{3}{11}\\ N=\frac{14}{11}\end{array}$

Vì $\frac{1}{3}<1<\frac{14}{11}$ nên M < 1 < N