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

Tìm x, biết a ) x + 2/5 = - 4/ 3

Giải thích

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

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

\(x = \frac{{ - 26}}{{15}}\)

Vậy \(x = \frac{{ - 26}}{{15}}\).

 

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

\(\frac{{ - 5}}{6} + \frac{1}{3} \cdot x = \frac{1}{4}\)

\(\frac{1}{3} \cdot x = \frac{1}{4} + \frac{5}{6}\)

\(\frac{1}{3} \cdot x = \frac{{13}}{{12}}\)

\(x = \frac{{13}}{4}\).

Vậy \(x = \frac{{13}}{4}\).

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

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

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

\[\left( {x + \frac{7}{6}} \right) \cdot \frac{5}{6} = \frac{{17}}{{24}}\]

\[x + \frac{7}{6} = \frac{{17}}{{24}}:\frac{5}{6}\]

\[x + \frac{7}{6} = \frac{{17}}{{20}}\]

\[x = \frac{{17}}{{20}} - \frac{7}{6}\]

\[x = \frac{{ - 19}}{{60}}\].

Vậy \[x = \frac{{ - 19}}{{60}}\].