Ta có: $\mathrm{x}:\mathrm{y}=\frac{\mathrm{a}}{\mathrm{b}}:\frac{\mathrm{c}}{\mathrm{d}}=\frac{\mathrm{a}}{\mathrm{b}}.\frac{\mathrm{d}}{\mathrm{c}}$

$$\begin{gathered} \left(-\frac{3}{4}+\frac{5}{13}-2\frac{1}{4}+\frac{8}{13}\right):\frac{2}{7} \\ =\left(-\frac{3}{4}+\frac{5}{13}-\frac{9}{4}+\frac{8}{13}\right):\frac{2}{7} \\ =\left(-\frac{3}{4}-\frac{9}{4}+\frac{5}{13}+\frac{8}{13}\right):\frac{2}{7} \\ =\left(-3+1\right):\frac{2}{7} \\ =\left(-2\right).\frac{7}{2}=-7 \end{gathered}$$

$$\begin{gathered} \frac{2}{3}-\frac{x}{15}=\frac{-2}{15} \\ \frac{x}{15}=\frac{2}{3}-\frac{-2}{15}=\frac{10}{15}+\frac{2}{15}=\frac{12}{15} \end{gathered}$$

$$\begin{gathered} \frac{2}{3}+\left(-\frac{3}{4}\right).\frac{4}{9} \\ =\frac{2}{3}+\left(\frac{-1}{3}\right) \\ =\frac{1}{3} \end{gathered}$$

Ta có: $\mathrm{x}.\mathrm{y}=\frac{\mathrm{a}}{\mathrm{b}}.\frac{\mathrm{c}}{\mathrm{d}}=\frac{\mathrm{a}.\mathrm{c}}{\mathrm{b}.\mathrm{d}}$.

Muốn nhân hai phân số, ta nhân các tử với nhau và nhân các mẫu với nhau.

$\frac{3}{2}.\frac{4}{7}=\frac{3.4}{2.7}=\frac{6}{7}>0$

Vậy kết quả của phép tính trên là một số nguyên dương.

Ta có:

$$\begin{gathered} \mathrm{A}=\frac{4}{5}.\frac{20}{8}.\left(-\frac{4}{3}\right) \\ =\frac{4.20.\left(-4\right)}{5.8.3}=\frac{-8}{3} \\ \mathrm{B}=\frac{-2}{11}.\frac{5}{18}.\left(\frac{-121}{25}\right) \\ =\frac{\left(-2\right).5.\left(-121\right)}{11.18.25}=\frac{11}{45} \end{gathered}$$

Do đó A < B.

$$\begin{gathered} \frac{1}{2}\mathrm{x}+\frac{3}{5}\mathrm{x}=\frac{-3}{65} \\ \left(\frac{1}{2}+\frac{3}{5}\right).\mathrm{x}=\frac{-3}{65} \\ \frac{11}{10}.\mathrm{x}=\frac{-3}{65} \\ \mathrm{x}=\frac{-3}{65}:\frac{11}{10} \\ \mathrm{x}=\frac{-6}{143} \end{gathered}$$

b) $\left(\frac{2}{9}\mathrm{x}-\frac{4}{9}\right)\left(\frac{1}{3}+\frac{-4}{7}:\mathrm{x}\right)=0$

TH1: $\frac{2}{9}\mathrm{x}-\frac{4}{9}=0$

$$\begin{gathered} \frac{2}{9}\mathrm{x}=\frac{4}{9} \\ \Rightarrow \mathrm{x}=2 \end{gathered}$$

TH2: $\frac{1}{3}+\frac{-4}{7}:\mathrm{x}=0$

$$\begin{gathered} \frac{-4}{7}:\mathrm{x}=\frac{-1}{3} \\ \mathrm{x}=\frac{12}{7} \end{gathered}$$

Suy ra $\mathrm{x}\in \left\{2; \frac{12}{7}\right\}$

Vậy có 2 giá trị của x thỏa mãn yêu cầu bài toán.

Ta có:

$$\begin{gathered} \mathrm{B}=\frac{{\displaystyle \frac{1}{6}-\frac{1}{39}+\frac{1}{51}}}{{\displaystyle \frac{1}{8}-\frac{1}{52}+\frac{1}{68}}} \\ =\frac{{\displaystyle \frac{1}{3}\left(\frac{1}{2}-\frac{1}{32}+\frac{1}{17}\right)}}{{\displaystyle \frac{1}{4}\left(\frac{1}{2}-\frac{1}{13}+\frac{1}{17}\right)}} \\ =\frac{1}{3}:\frac{1}{4}=\frac{4}{3} \end{gathered}$$

$$\begin{gathered} \mathrm{x}=\frac{7}{4}.\left(-\frac{2}{5}\right) \\ =\frac{7.\left(-2\right)}{4.5} \\ =\frac{-7}{10}; \\ \mathrm{y}=\frac{5}{11}:\frac{15}{22} \\ =\frac{5}{11}.\frac{22}{15} \\ =\frac{5.22}{11.15} \\ \mathrm{y}=\frac{5.2}{3}=\frac{10}{3} \\ 2\mathrm{x}+\mathrm{y}=2.\left(\frac{-7}{10}\right)+\frac{10}{3} \\ 2\mathrm{x}+\mathrm{y}=\frac{-7}{5}+\frac{10}{3} \\ 2\mathrm{x}+\mathrm{y}=\frac{-21}{15}+\frac{50}{15} \\ 2\mathrm{x}+\mathrm{y}=\frac{29}{15} \end{gathered}$$

$$\begin{gathered} \frac{\mathrm{x}+2}{338}+1+\frac{\mathrm{x}+3}{337}+1+\frac{\mathrm{x}+4}{336}+1+\frac{\mathrm{x}+5}{335}+1+\frac{\mathrm{x}+360}{5}-4=0 \\ \iff \frac{\mathrm{x}+340}{338}+\frac{\mathrm{x}+340}{337}+\frac{\mathrm{x}+340}{336}+\frac{\mathrm{x}+340}{335}+\frac{\mathrm{x}+340}{5}=0 \\ \iff \left(\mathrm{x}+340\right)\left(\frac{1}{338}+\frac{1}{337}+\frac{1}{336}+\frac{1}{335}+\frac{1}{5}\right)=0 \end{gathered}$$

Mà $\frac{1}{338}+\frac{1}{337}+\frac{1}{336}+\frac{1}{5}\ne 0$

Suy ra x = -340.

$$\begin{gathered} \mathrm{D}=\left[\frac{-54}{64}-\left(\frac{1}{9}:\frac{8}{27}\right):\frac{-1}{3}\right]:\frac{-81}{128} \\ \mathrm{D}=\left[-\frac{27}{32}-\frac{3}{8}.\frac{3}{-1}\right].\frac{128}{-81} \\ \mathrm{D}=\left[\frac{-27}{32}+\frac{9}{8}\right].\frac{128}{-81} \\ \mathrm{D}=\left[\frac{-27+36}{32}.\frac{128}{-81}\right] \\ \mathrm{D}=\frac{9}{32}.\frac{128}{-81}=\frac{-4}{9} \end{gathered}$$

Suy ra: $\mathrm{D}+\frac{13}{9}=\frac{-4}{9}+\frac{13}{9}=\frac{9}{9}=1$

Ta có:

$$\begin{gathered} \mathrm{A}=\frac{5-{\displaystyle \frac{5}{13}}+{\displaystyle \frac{5}{19}}-{\displaystyle \frac{5}{27}}}{11-{\displaystyle \frac{11}{3}}+{\displaystyle \frac{11}{19}}-{\displaystyle \frac{11}{27}}}+\frac{{\displaystyle \frac{6}{101}+\frac{6}{123}-\frac{6}{124}}}{{\displaystyle \frac{11}{101}+\frac{11}{123}-\frac{11}{134}}} \\ =\frac{5\left(1-{\displaystyle \frac{1}{13}}+{\displaystyle \frac{1}{19}}-{\displaystyle \frac{1}{27}}\right)}{11\left(1-{\displaystyle \frac{1}{3}}+{\displaystyle \frac{1}{19}}-{\displaystyle \frac{1}{27}}\right)}+\frac{6\left({\displaystyle \frac{1}{101}}+{\displaystyle \frac{1}{123}}-{\displaystyle \frac{1}{134}}\right)}{11.\left({\displaystyle \frac{1}{101}}+{\displaystyle \frac{1}{123}}-{\displaystyle \frac{1}{134}}\right)} \\ \Rightarrow \mathrm{A}=\frac{5}{11}+\frac{6}{11}=1 \end{gathered}$$