Giải bài 7 trang 43 SGK Toán 6 Cánh diều

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)

\(\begin{array}{l}\frac{{17}}{8}:\left( {\frac{{27}}{8} + \frac{{11}}{2}} \right)\\ = \frac{{17}}{8}:\left( {\frac{{27}}{8} + \frac{{44}}{8}} \right)\\ = \frac{{17}}{8}:\frac{{71}}{8}\\ = \frac{{17}}{8}.\frac{8}{{71}}\\ = \frac{{17}}{{71}}\end{array}\)

b)

\(\begin{array}{l}\frac{{28}}{{15}}.\frac{1}{{{4^2}}}.3 + \left( {\frac{8}{{15}} - \frac{{69}}{{60}}.\frac{5}{{23}}} \right):\frac{{51}}{{54}}\\ = \frac{{28.1.3}}{{{{15.4}^2}}} + \left( {\frac{8}{{15}} - \frac{{23.3}}{{4.3.5}}.\frac{5}{{23}}} \right).\frac{{54}}{{51}}\\ = \frac{{7.4.1.3}}{{3.5.4.4}} + \left( {\frac{8}{{15}} - \frac{1}{4}} \right).\frac{{54}}{{51}}\\ = \frac{7}{{20}} + \left( {\frac{{32}}{{60}} - \frac{{15}}{{60}}} \right).\frac{{54}}{{51}}\\ = \frac{7}{{20}} + \frac{{17}}{{60}}.\frac{{54}}{{51}}\\ = \frac{7}{{20}} + \frac{{17}}{{6.10}}.\frac{{6.3.3}}{{17.3}}\\ = \frac{7}{{20}} + \frac{3}{{10}}\\ = \frac{7}{{20}} + \frac{6}{{20}}\\ = \frac{{13}}{{20}}\end{array}\)