Tính: a) $\frac{7}{8}+\frac{7}{8}:\frac{1}{8}-\frac{1}{2};$              b) $\frac{6}{11}+\frac{11}{3}.\frac{3}{22}.$

Giải

$$\begin{gathered} \mathrm{a}) \frac{7}{8}+\frac{7}{8}:\frac{1}{8}-\frac{1}{2} \\ =\frac{7}{8}.1+\frac{7}{8}.8-\frac{1}{2} \\ =\frac{7}{8}(1+8)-\frac{1}{2} (\mathrm{tính} \mathrm{chất} \mathrm{phân} \mathrm{phối} \mathrm{của} \mathrm{phép} \mathrm{nhân} \mathrm{đối} \mathrm{với} \mathrm{phép} \mathrm{cộng}) \\ =\frac{7}{8}.9-\frac{1}{2} \\ =\frac{7.9}{8}-\frac{1.4}{2.4} \\ =\frac{63}{8}-\frac{4}{8} \\ =\frac{63-4}{8} \\ =\frac{59}{8} \end{gathered}$$

b)

$$\begin{gathered} \frac{6}{11}+\frac{11}{3}.\frac{3}{22} \\ =\frac{6}{11}+\frac{11.3}{3.22} \\ =\frac{6}{11}+\frac{11}{22} \\ =\frac{6.2}{11.2}+\frac{11}{22} \\ =\frac{12}{22}+\frac{11}{22} \\ =\frac{12+11}{22} \\ =\frac{23}{22} \end{gathered}$$