Giải bài 7 trang 43

Đề bài

Tính:

a) $\frac{17}{8}:\left(\frac{27}{8}+\frac{-11}{2}\right)$;

b) $\frac{28}{15}.\frac{1}{4^2}.3+\left(\frac{8}{15}-\frac{69}{60}.\frac{5}{23}\right):\frac{-51}{54}$.

Lời giải chi tiết

a)

$\begin{array}{c}\frac{17}{8}:\left(\frac{27}{8}+\frac{-11}{2}\right)\\ =\frac{17}{8}:\left(\frac{27}{8}+\frac{-44}{8}\right)\\ =\frac{17}{8}:\frac{-17}{8}\\ =\frac{17}{8}.\frac{-8}{17}\\ =-1\end{array}$

b)

$\begin{array}{c}\frac{28}{15}.\frac{1}{4^2}.3+\left(\frac{8}{15}-\frac{69}{60}.\frac{5}{23}\right):\frac{-51}{54}\\ =\frac{\mathrm{28.1.3}}{{15.4}^2}+\left(\frac{8}{15}-\frac{23.3}{\mathrm{4.3.5}}.\frac{5}{23}\right).\frac{-54}{51}\\ =\frac{\mathrm{7.4.1.3}}{\mathrm{3.5.4.4}}+\left(\frac{8}{15}-\frac{1}{4}\right).\frac{-54}{51}\\ =\frac{7}{20}+\left(\frac{32}{60}-\frac{15}{60}\right).\frac{-54}{51}\\ =\frac{7}{20}+\frac{17}{60}.\frac{-54}{51}\\ =\frac{7}{20}+\frac{17}{6.10}.\frac{-\mathrm{6.3.3}}{17.3}\\ =\frac{7}{20}+\frac{-3}{10}\\ =\frac{7}{20}+\frac{-6}{20}\\ =\frac{1}{20}\end{array}$