Giải bài 5 trang 27 SGK Toán 7 tập 1 - Chân trời sáng tạo

Đề bài

Tìm x, biết:

a)$- \frac{3}{5}.x = \frac{{12}}{{25}};$

b)$\frac{3}{5}x - \frac{3}{4} =  - 1\frac{1}{2};$

c)$\frac{2}{5} + \frac{3}{5}:x = 0,5;$

d)$\frac{3}{4} - \left( {x - \frac{1}{2}} \right) = 1\frac{2}{3}$

e)$2\frac{2}{{15}}:\left( {\frac{1}{3} - 5x} \right) =  - 2\frac{2}{5}$

g)${x^2} + \frac{1}{9} = \frac{5}{3}:3.$

Lời giải chi tiết

a)

$\begin{array}{l} - \frac{3}{5}.x = \frac{{12}}{{25}}\\x = \frac{{12}}{{25}}:\frac{{ - 3}}{5}\\x = \frac{{12}}{{25}}.\frac{{ - 5}}{3}\\x = \frac{{ - 4}}{5}\end{array}$

Vậy $x = \frac{{ - 4}}{5}$

b)

$\begin{array}{l}\frac{3}{5}x - \frac{3}{4} =  - 1\frac{1}{2};\\\frac{3}{5}x = \frac{{ - 3}}{2} + \frac{3}{4}\\\frac{3}{5}x = \frac{{ - 3}}{4}\\x = \frac{{ - 3}}{4}:\frac{3}{5}\\x = \frac{{ - 3}}{4}.\frac{5}{3}\\x = \frac{{ - 5}}{4}\end{array}$

Vậy $x = \frac{{ - 5}}{4}$.

c)

$\begin{array}{l}\frac{2}{5} + \frac{3}{5}:x = 0,5\\\frac{3}{5}:x = \frac{1}{2} - \frac{2}{5}\\\frac{3}{5}:x = \frac{1}{{10}}\\x = \frac{3}{5}:\frac{1}{{10}}\\x = \frac{3}{5}.10\\x = 6\end{array}$

Vậy $x = 6$.

d)

$\begin{array}{l}\frac{3}{4} - \left( {x - \frac{1}{2}} \right) = 1\frac{2}{3}\\x - \frac{1}{2} = \frac{3}{4} - \frac{5}{3}\\x - \frac{1}{2} = \frac{{ - 11}}{{12}}\\x = \frac{{ - 11}}{{12}} + \frac{1}{2}\\x = \frac{{ - 5}}{{12}}\end{array}$

Vậy $x = \frac{{ - 5}}{{12}}$.

e)

$\begin{array}{l}2\frac{2}{{15}}:\left( {\frac{1}{3} - 5x} \right) =  - 2\frac{2}{5}\\\frac{{32}}{{15}}:\left( {\frac{1}{3} - 5x} \right) =  - \frac{{12}}{5}\\\frac{1}{3} - 5x = \frac{{32}}{{15}}:\frac{{ - 12}}{5}\\\frac{1}{3} - 5x = \frac{{32}}{{15}}.\frac{{ - 5}}{12}\\\frac{1}{3} - 5x = \frac{{ - 8}}{9}\\5x = \frac{1}{3} + \frac{8}{9}\\5x = \frac{{11}}{9}\\x = \frac{{11}}{9}:5\\x = \frac{{11}}{{45}}\end{array}$

Vậy $x = \frac{{11}}{{45}}$.

g)

${x^2} + \frac{1}{9} = \frac{5}{3}:3\\{x^2} + \frac{1}{9} = \frac{5}{9}\\{x^2} = \frac{5}{9} - \frac{1}{9}\\{x^2} = \frac{4}{9}\\{x^2} = (\frac{2}{3})^2\\x = \frac{2}{3}\,\ hoặc \,\ x = \frac{-2}{3}$

Vậy $x \in \{\frac{2}{3};\frac{-2}{3}\}$.