Giải bài 7 trang 43 SGK Toán 6 Cánh diều — Không quảng cáo

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)

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

b)

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