Ta đổi các số thập phân trong biểu thức trên ra phân số như sau:

Khi đó, ta có thể tính giá trị của biểu thức trên như sau:

$$\begin{gathered} 0,5+4,5:3-\frac{3}{16}.\frac{4}{3} \\ =\frac{1}{2}+\frac{9}{2}:3-\frac{3}{16}.\frac{4}{3} \\ =\frac{1}{2}+\frac{9}{2}.\frac{1}{3}-\frac{3}{16}.\frac{4}{3} \\ =\frac{1}{2}+\frac{3}{2}-\frac{1}{4} \\ =2-\frac{1}{4} \\ =\frac{8}{4}-\frac{1}{4}=\frac{7}{4} \end{gathered}$$

$$\begin{gathered} a) 0,2+2,5:\frac{7}{2}=\frac{1}{5}+\frac{5}{2}:\frac{7}{2} \\ =\frac{1}{5}+\frac{5}{2}.\frac{2}{7}=\frac{1}{5}+\frac{5}{7} \\ =\frac{7}{35}+\frac{25}{35}=\frac{32}{35} \\ b) 9.{\left(\frac{-1}{3}\right)}^2-{(-0,1)}^3:\frac{2}{15} \\ =9.\frac{1}{9}-\left(-\frac{1}{1000}\right):\frac{2}{15} \\ =1+\frac{1}{1000}:\frac{2}{15} \\ =1+\frac{{\displaystyle 1}}{{\displaystyle 1000}}.\frac{{\displaystyle 15}}{{\displaystyle 2}} \\ =\frac{{\displaystyle 400}}{{\displaystyle 400}}+\frac{{\displaystyle 3}}{{\displaystyle 400}}=\frac{{\displaystyle 403}}{{\displaystyle 400}} \end{gathered}$$

$$\begin{gathered} a) \left(0,25-\frac{5}{6}\right).1,6+\frac{-1}{3} \\ =\left(\frac{1}{4}-\frac{5}{6}\right).\frac{8}{5}+\frac{-1}{3} \\ =\left(\frac{3}{12}-\frac{10}{12}\right).\frac{8}{5}+\frac{-1}{3} \\ =\frac{-7}{12}.\frac{8}{5}+\frac{-1}{3} \\ =\frac{-14}{15}+\frac{-1}{3} \\ =\frac{-14}{15}+\frac{-5}{15} \\ =\frac{-19}{15} \\ b) 3-2.\left[0,5+\left(0,25-\frac{1}{6}\right)\right] \\ =3-2.\left[\frac{1}{2}+\left(\frac{1}{4}-\frac{1}{6}\right)\right] \\ =3-2.\left[\frac{1}{2}+\left(\frac{3}{12}-\frac{2}{12}\right)\right] \\ =3-2.\left(\frac{1}{2}+\frac{1}{12}\right) \\ =3-2.\left(\frac{6}{12}+\frac{1}{12}\right) \\ =3-2.\frac{7}{12}=3-\frac{7}{6} \\ =\frac{18}{6}-\frac{7}{6}=\frac{11}{6} \end{gathered}$$

$$\begin{gathered} a) 1,8-\left(\frac{3}{7}-0,2\right) \\ =\frac{9}{5}-\left(\frac{3}{7}-\frac{1}{5}\right) \\ =\frac{9}{5}-\frac{3}{7}+\frac{1}{5} \\ =\frac{9}{5}+\frac{1}{5}-\frac{3}{7} \\ =2-\frac{3}{7} \\ =\frac{14}{7}-\frac{3}{7}=\frac{11}{7} \\ b) 12,5-\frac{{\displaystyle 16}}{{\displaystyle 13}}+\frac{{\displaystyle 3}}{{\displaystyle 13}} \\ =12,5-\left(\frac{{\displaystyle 16}}{{\displaystyle 13}}-\frac{{\displaystyle 3}}{{\displaystyle 13}}\right)= 12,5 – 1 = 11,5. \end{gathered}$$

$$\begin{gathered} a) \left(-\frac{5}{6}\right)-(-1,8)+\left(-\frac{1}{6}\right)-0,8 \\ =\left(-\frac{5}{6}\right)+\left(-\frac{1}{6}\right)-(-1,8)-0,8 \\ =-\left(\frac{5}{6}+\frac{1}{6}\right)+(1,8-0,8)= (- 1) + 1 = 0. \\ b) \left(-\frac{{\displaystyle 9}}{{\displaystyle 7}}\right)+(-1,23)-\left(-\frac{{\displaystyle 2}}{{\displaystyle 7}}\right)-0,77 \\ =\left(-\frac{{\displaystyle 9}}{{\displaystyle 7}}\right)-1,23+\frac{{\displaystyle 2}}{{\displaystyle 7}}-0,77 \\ =\left(-\frac{{\displaystyle 9}}{{\displaystyle 7}}\right)+\frac{{\displaystyle 2}}{{\displaystyle 7}}-1,23-0,77 \\ =-\left(\frac{{\displaystyle 9}}{{\displaystyle 7}}-\frac{{\displaystyle 2}}{{\displaystyle 7}}\right)-(1,23+0,77) \\ = (- 1) – 2 = – (1 + 2) = – 3. \end{gathered}$$

$$\begin{gathered} a) \frac{1}{9}-0,3.\frac{5}{9}+\frac{1}{3} \\ =\frac{1}{9}-\frac{3}{10}.\frac{5}{9}+\frac{1}{3} \\ =\frac{1}{9}-\frac{1}{6}+\frac{1}{3} \\ =\frac{2}{18}-\frac{3}{18}+\frac{6}{18}=\frac{5}{18} \\ b) {\left(\frac{{\displaystyle -2}}{{\displaystyle 3}}\right)}^2+\frac{{\displaystyle 1}}{{\displaystyle 6}}-{(-0,5)}^3 \\ =\frac{{(-2)}^2}{3^2}+\frac{1}{6}-{\left(\frac{-1}{2}\right)}^3 \\ =\frac{{\displaystyle 4}}{{\displaystyle 9}}+\frac{{\displaystyle 1}}{{\displaystyle 6}}-\frac{{\displaystyle {(-1)}^3}}{{\displaystyle 2^3}} \\ =\frac{{\displaystyle 4}}{{\displaystyle 9}}+\frac{{\displaystyle 1}}{{\displaystyle 6}}-\frac{{\displaystyle -1}}{{\displaystyle 8}} \\ =\frac{{\displaystyle 4}}{{\displaystyle 9}}+\frac{{\displaystyle 1}}{{\displaystyle 6}}+\frac{{\displaystyle 1}}{{\displaystyle 8}} \\ =\frac{{\displaystyle 32}}{{\displaystyle 72}}+\frac{{\displaystyle 12}}{{\displaystyle 72}}+\frac{{\displaystyle 9}}{{\displaystyle 72}}=\frac{{\displaystyle 53}}{{\displaystyle 72}} \end{gathered}$$

$$\begin{gathered} a) \left(\frac{4}{5}-1\right):\frac{3}{5}-\frac{2}{3}.0,5 \\ =\left(\frac{-1}{5}\right):\frac{3}{5}-\frac{2}{3}.\frac{1}{2} \\ =\left(\frac{-1}{5}\right).\frac{5}{3}-\frac{2}{3}.\frac{1}{2} \\ =\left(\frac{-1}{3}\right)-\frac{1}{3} \\ =-\left(\frac{1}{3}+\frac{1}{3}\right)=-\frac{2}{3} \end{gathered}$$

$$\begin{gathered} b) 1-{\left(\frac{5}{9}-\frac{2}{3}\right)}^2:\frac{4}{27}=1-{\left(\frac{5}{9}-\frac{6}{9}\right)}^2:\frac{4}{27} \\ =1-{\left(\frac{-1}{9}\right)}^2.\frac{27}{4}=1-\frac{{(-1)}^2}{9^2}.\frac{27}{4} \\ =1-\frac{1}{81}.\frac{27}{4}=1-\frac{1}{12}=\frac{11}{12} \end{gathered}$$

$$\begin{gathered} c) \left[\left(\frac{3}{8}-\frac{5}{12}\right).6+\frac{1}{3}\right].4=\left[\left(\frac{9}{24}-\frac{10}{24}\right).6+\frac{1}{3}\right].4 \\ =\left(\frac{-1}{24}.6+\frac{1}{3}\right).4=\left(\frac{-1}{=\frac{1}{12}.4=\frac{1}{3}4}+\frac{1}{3}\right).4 \\ =\left(\frac{-3}{}+\frac{4}{12}\right).4 \end{gathered}$$

$$\begin{gathered} d) 0,8:\left\{0,2-7.\left[\frac{1}{6}+\left(\frac{5}{21}-\frac{5}{14}\right)\right]\right\} \\ =\frac{4}{5}:\left\{\frac{1}{5}-7.\left[\frac{1}{6}+\left(\frac{10}{42}-\frac{15}{42}\right)\right]\right\} \\ =\frac{4}{5}:\left\{\frac{1}{5}-7.\left[\frac{1}{6}+\frac{-5}{42}\right]\right\} \\ =\frac{4}{5}:\left\{\frac{1}{5}-7.\left[\frac{7}{42}+\frac{-5}{42}\right]\right\} \\ =\frac{4}{5}:\left\{\frac{1}{5}-7.\frac{1}{21}\right\}=\frac{4}{5}:\left\{\frac{1}{5}-\frac{1}{3}\right\} \\ =\frac{4}{5}:\left\{\frac{3}{15}-\frac{5}{15}\right\}=\frac{4}{5}:\frac{-2}{15}=\frac{4}{5}.\frac{-15}{2}=-6 \end{gathered}$$

$$\begin{gathered} a) Ta có \left(3,9+\frac{1}{2}.2,6\right):13 \\ =3,9:13+\frac{1}{2}.2,6:13=0,3+\frac{1}{2}.2,6:13 \\ Vì ,3 \ne 3,9 nên \left(3,9+\frac{1}{2}.2,6\right):13\ne 3,9+\frac{1}{2}.2,6:13 \\ Vậy \left(3,9+\frac{1}{2}.2,6\right):13\boxed{\ne }3,9+\frac{1}{2}.2,6:13 \end{gathered}$$

$$\begin{gathered} b) \frac{5}{4}-{\left(\frac{1}{4}\right)}^2=\frac{5}{4}-\frac{1^2}{4^2}=\frac{5}{4}-\frac{1}{16}=\frac{20}{16}-\frac{1}{16}=\frac{19}{16} \\ {\left(\frac{5}{4}-\frac{1}{4}\right)}^2=1^2=1 \\ Vì \frac{19}{16}\ne 1 nên \frac{5}{4}-{\left(\frac{1}{4}\right)}^2\ne {\left(\frac{5}{4}-\frac{1}{4}\right)}^2 \\ \text{Vậy }\frac{5}{4}-{\left(\frac{1}{4}\right)}^2\boxed{\ne }{\left(\frac{5}{4}-\frac{1}{4}\right)}^2 \end{gathered}$$

$$\begin{gathered} c) Ta có \frac{28}{9}.0,7+\frac{28}{9}.0,5=\frac{28}{9}.(0,7+0,5) \\ \frac{28}{9}.0,7+\frac{28}{9}.0,5\boxed{=}\frac{28}{9}.(0,7+0,5) \end{gathered}$$

$$\begin{gathered} d) Ta có \frac{36}{13}:4+\frac{36}{13}:9=\frac{36}{13}.\frac{1}{4}+\frac{36}{13}.\frac{1}{9} \\ =\frac{36}{13}.\left(\frac{1}{4}+\frac{1}{9}\right)=\frac{36}{13}.\left(\frac{9}{36}+\frac{4}{36}\right)=\frac{36}{13}.\frac{13}{36} \\ \frac{36}{13}:(4+9)=\frac{36}{13}:13=\frac{36}{13}.\frac{1}{13}=\frac{36}{13}.\frac{3}{36} \\ \text{Vì }\frac{13}{36}\ne \frac{3}{36} nên \frac{36}{13}.\frac{13}{36}\ne \frac{36}{13}.\frac{3}{36} \\ \text{Vậy }\frac{36}{13}.\frac{13}{36}\ne \frac{36}{13}.\frac{3}{36} \end{gathered}$$

$$\begin{gathered} a) \frac{4}{15}-\left(2,9-\frac{11}{15}\right)=\frac{4}{15}-2,9+\frac{11}{15} \\ =\frac{4}{15}+\frac{11}{15}-2,9= 1 – 2,9 = – 1,9. \\ b) (-36,75)+\left(\frac{37}{10}-63,25\right)-(-6,3) \\ =(-36,75)+\frac{37}{10}-63,25+6,3 \\ = (– 36,75) – 63,25 + 3,7 + 6,3 \\ = – (36,75 + 63,25) + (3,7 + 6,3) \\ = (– 100) + 10 = – 90. \\ c) 6,5+\left(-\frac{10}{17}\right)-\left(-\frac{7}{2}\right)-\frac{7}{17} \\ =6,5-\frac{10}{17}+3,5-\frac{7}{17}=6,5+3,5-\frac{10}{17}-\frac{7}{17} \\ =(6,5+3,5)-\left(\frac{10}{17}+\frac{7}{17}\right)\text{= 10 – 1 = 9.} \end{gathered}$$

$$\begin{gathered} d) (-39,1).\frac{13}{25}-60,9.\frac{13}{25} \\ =-\frac{13}{25}.(39,1+60,9)=-\frac{13}{25}.100=-52 \end{gathered}$$

Cứ $\frac{1}{4}$m trồng một khóm hoa, nghĩa là mỗi khóm hoa cách nhau 0,25 m.