Bộ 5 đề thi giữa kì 1 Toán 7 Cánh diều (2022-2023) có đáp án - Đề 1

Tìm x, biết x + 2/3 = -4/5

Giải thích

a) \(x + \frac{2}{3} = \frac{{ - 4}}{5}\)

\(x = \frac{{ - 4}}{5} - \frac{2}{3}\)

\(x = \frac{{ - 22}}{{15}}\)

Vậy \(x = \frac{{ - 22}}{{15}}\)

b) \(\frac{{ - 3}}{4} + \frac{1}{3} \cdot x = \frac{1}{6}\)

\(\frac{1}{3} \cdot x = \frac{1}{6} + \frac{3}{4}\)

\(\frac{1}{3} \cdot x = \frac{{11}}{{12}}\)

\(x = \frac{{11}}{4}\)

Vậy \(x = \frac{{11}}{4}\)

c) \(\frac{7}{{12}} - \left( {x + \frac{1}{6}} \right) \cdot \frac{6}{5} = {\left( {\frac{{ - 1}}{2}} \right)^3}\)

\[\frac{7}{{12}} - \left( {x + \frac{1}{6}} \right) \cdot \frac{5}{6} = \frac{{ - 1}}{8}\]

\[\left( {x + \frac{1}{6}} \right) \cdot \frac{5}{6} = \frac{7}{{12}} - \frac{{ - 1}}{8}\]

\[\left( {x + \frac{1}{6}} \right) \cdot \frac{5}{6} = \frac{{17}}{{24}}\]

\[x + \frac{1}{6} = \frac{{17}}{{24}}:\frac{5}{6}\]

\[x + \frac{1}{6} = \frac{{17}}{{20}}\]

\[x = \frac{{17}}{{20}} - \frac{1}{6}\]

\[x = \frac{{41}}{{60}}\]

Vậy \[x = \frac{{41}}{{60}}\]