Giải bài 1.8 trang 13 SGK Toán 7 tập 1 – Kết nối tri thức

Đề bài

Tính giá trị của các biểu thức sau:

\(\begin{array}{l}a)(8 + 2\frac{1}{3} – \frac{3}{5}) – (5 + 0,4) – (3\frac{1}{3} – 2)\\b)(7 – \frac{1}{2} – \frac{3}{4}):(5 – \frac{1}{4} – \frac{5}{8})\end{array}\)

Lời giải chi tiết

a) Cách 1:

\(\begin{array}{l}(8 + 2\frac{1}{3} – \frac{3}{5}) – (5 + 0,4) – (3\frac{1}{3} – 2)\\ = (8 + \frac{7}{3} – \frac{3}{5}) – (5 + \frac{4}{{10}}) – (\frac{{10}}{3} – 2)\\ = 8 + \frac{7}{3} – \frac{3}{5} – 5 – \frac{2}{5} – \frac{{10}}{3} + 2\\ = (8 – 5 + 2) + (\frac{7}{3} – \frac{{10}}{3}) – (\frac{3}{5} + \frac{2}{5})\\ = 5 + \frac{{ – 3}}{3} – \frac{5}{5}\\ = 5 + ( – 1) – 1\\ = 3\end{array}\)

Cách 2:

\(\begin{array}{l}(8 + 2\frac{1}{3} – \frac{3}{5}) – (5 + 0,4) – (3\frac{1}{3} – 2)\\ = (8 + \frac{7}{3} – \frac{3}{5}) – (5 + \frac{4}{{10}}) – (\frac{{10}}{3} – 2)\\ = (\frac{{120}}{{15}} + \frac{{35}}{{15}} – \frac{9}{{15}}) – (\frac{{25}}{5} + \frac{2}{5}) – (\frac{{10}}{3} – \frac{6}{3})\\ = \frac{{146}}{{15}} – \frac{{27}}{5} – \frac{4}{3}\\ = \frac{{146}}{{15}} – \frac{{81}}{{15}} – \frac{{20}}{{15}}\\ = \frac{{45}}{{15}}\\ = 3\end{array}\)

b)

\(\begin{array}{l}(7 – \frac{1}{2} – \frac{3}{4}):(5 – \frac{1}{4} – \frac{5}{8})\\ = (\frac{{28}}{4} – \frac{2}{4} – \frac{3}{4}):(\frac{{40}}{8} – \frac{2}{8} – \frac{5}{8})\\ = \frac{{23}}{4}:\frac{{33}}{8}\\ = \frac{{23}}{4}.\frac{8}{{33}}\\ = \frac{{46}}{{33}}\end{array}\)