Giải bài 4 trang 25 SGK Toán 7 tập 1 – Chân trời sáng tạo

Đề bài

Tìm x, biết:

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

b)\(\frac{3}{7} – x = \frac{2}{5};\)

c)\(\frac{4}{9} – \frac{2}{3}x = \frac{1}{3};\)                   

d)\(\frac{3}{{10}}x – 1\frac{1}{2} = \left( {\frac{{ – 2}}{7}} \right):\frac{5}{{14}}\)

Lời giải chi tiết

a)

\(\begin{array}{l}x + \frac{3}{5} = \frac{2}{3}\\x = \frac{2}{3} – \frac{3}{5}\\x = \frac{{10}}{{15}} – \frac{9}{{15}}\\x = \frac{1}{{15}}\end{array}\)                      

Vậy \(x = \frac{1}{{15}}\).

b)

\(\begin{array}{l}\frac{3}{7} – x = \frac{2}{5}\\x = \frac{3}{7} – \frac{2}{5}\\x = \frac{{15}}{{35}} – \frac{{14}}{{35}}\\x = \frac{1}{{35}}\end{array}\)

Vậy \(x = \frac{1}{{35}}\).

c)

\(\begin{array}{l}\frac{4}{9} – \frac{2}{3}x = \frac{1}{3}\\\frac{2}{3}x = \frac{4}{9} – \frac{1}{3}\\\frac{2}{3}x = \frac{4}{9} – \frac{3}{9}\\\frac{2}{3}x = \frac{1}{9}\\x = \frac{1}{9}:\frac{2}{3}\\x = \frac{1}{9}.\frac{3}{2}\\x = \frac{1}{6}\end{array}\)                   

Vậy \(x = \frac{1}{6}\).

d)

\(\begin{array}{l}\frac{3}{{10}}x – 1\frac{1}{2} = \left( {\frac{{ – 2}}{7}} \right):\frac{5}{{14}}\\\frac{3}{{10}}x – \frac{3}{2} = \left( {\frac{{ – 2}}{7}} \right).\frac{{14}}{5}\\\frac{3}{{10}}x – \frac{3}{2} = \frac{{ – 4}}{5}\\\frac{3}{{10}}x = \frac{{ – 4}}{5} + \frac{3}{2}\\\frac{3}{{10}}x = \frac{{ – 8}}{{10}} + \frac{{15}}{{10}}\\\frac{3}{{10}}x = \frac{7}{{10}}\\x = \frac{7}{{10}}:\frac{3}{{10}}\\x = \frac{7}{3}\end{array}\)

Vậy \(x = \frac{7}{3}\).