Giải Bài 5 trang 11 sách bài tập toán 7 tập 1 - Chân trời sáng tạo

Đề bài

Tìm x, biết:

a) \(x.\dfrac{{21}}{{25}} = \dfrac{{ - 7}}{{10}}\)

b) \(\dfrac{{ - 3}}{{20}}x = \dfrac{5}{{12}} - \dfrac{2}{5}\)

c) \(\dfrac{2}{9}:x = \dfrac{1}{{18}}:0,375\)

d) \(\left( { - \dfrac{{11}}{{12}}} \right):2x = \dfrac{5}{2} + \dfrac{1}{4}\)

Lời giải chi tiết

\(\begin{array}{l}a)x.\dfrac{{21}}{{25}} = \dfrac{{ - 7}}{{10}}\\ \Leftrightarrow x = \dfrac{{ - 7}}{{10}}:\dfrac{{21}}{{25}}\\ \Leftrightarrow x = \dfrac{{ - 7}}{{10}}.\dfrac{{25}}{{21}}\\ \Leftrightarrow x = \dfrac{{ - 5}}{6}\end{array}\)

Vậy \(x = \dfrac{{ - 5}}{6}\)

\(\begin{array}{l}b)\dfrac{{ - 3}}{{20}}x = \dfrac{5}{{12}} - \dfrac{2}{5}\\ \Leftrightarrow \dfrac{{ - 3}}{{20}}x = \dfrac{{25}}{{60}} - \dfrac{{24}}{{60}}\\ \Leftrightarrow \dfrac{{ - 3}}{{20}}x = \dfrac{1}{{60}}\\ \Leftrightarrow x = \dfrac{1}{{60}}:\dfrac{{ - 3}}{{20}}\\ \Leftrightarrow x = \dfrac{1}{{60}}.\dfrac{{ - 20}}{3}\\ \Leftrightarrow x = \dfrac{{ - 1}}{9}\end{array}\)

Vậy \(x = \dfrac{{ - 1}}{9}\)

\(\begin{array}{l}c)\dfrac{2}{9}:x = \dfrac{1}{{18}}:0,375\\ \Leftrightarrow \dfrac{2}{9}:x = \dfrac{1}{{18}}:\dfrac{3}{8}\\ \Leftrightarrow \dfrac{2}{9}:x = \dfrac{1}{{18}}.\dfrac{8}{3}\\ \Leftrightarrow \dfrac{2}{9}:x = \dfrac{4}{{27}}\\ \Leftrightarrow x = \dfrac{2}{9}:\dfrac{4}{{27}}\\\Leftrightarrow x = \dfrac{2}{9}.\dfrac{27}{{4}}\\ \Leftrightarrow x = \dfrac{3}{2}\end{array}\)

Vậy \(x = \dfrac{3}{2}\)

\(\begin{array}{l}d)\left( { - \dfrac{{11}}{{12}}} \right):2x = \dfrac{5}{2} + \dfrac{1}{4}\\ \Leftrightarrow \left( { - \dfrac{{11}}{{12}}} \right):2x = \dfrac{{10}}{4} + \dfrac{1}{4}\\ \Leftrightarrow \left( { - \dfrac{{11}}{{12}}} \right):2x = \dfrac{{11}}{4}\\ \Leftrightarrow \left( { - \dfrac{{11}}{{12}}} \right):\dfrac{{11}}{4} = 2x\\ \Leftrightarrow \dfrac{{ - 11}}{{12}}.\dfrac{4}{{11}} = 2x\\ \Leftrightarrow \dfrac{{ - 1}}{3} = 2x\end{array}\)

\(\begin{array}{l} \Leftrightarrow x = \dfrac{{ - 1}}{3}:2\\\Leftrightarrow x = \dfrac{{ - 1}}{3}.\dfrac{1}{2}\\ \Leftrightarrow x = \dfrac{{ - 1}}{6}\end{array}\)

Vậy \(x = \dfrac{{ - 1}}{6}\)