Đề bài - giải bài 7 trang 43 sgk toán 6 cánh diều

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

Đề bài

Tính:

a] \[\frac{{17}}{8}:\left[ {\frac{{27}}{8} + \frac{{11}}{2}} \right]\];

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

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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]

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

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