Bài 9 trang 10 SGK Toán 7 tập 1

Đề bài

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

b) x - \(\frac{2}{5} = \frac{5}{7}\)

c) -x - \(\frac{2}{3}\) = \(- \frac{6}{7}\)

d) \(\frac{4}{7} - x = \frac{1}{3}\)

Hướng dẫn giải

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

     x = \(\frac{3}{4} - \frac{1}{3} = \frac{9}{12} - \frac{4}{12} = \frac{5}{12}\)

b)  x - \(\frac{2}{5} = \frac{5}{7}\)

     x = \(\frac{5}{7} + \frac{2}{5} = \frac{25}{35} + \frac{14}{35}= \frac{39}{35} = 1\frac{4}{35}\)

c) -x - \(\frac{2}{3}\) = \(- \frac{6}{7}\)

    \(\frac{-2}{3} + \frac{6}{7} = x \)

    \(x = -\frac{14}{21} + \frac{18}{21} = \frac{4}{21}\)

d) \(\frac{4}{7} - x = \frac{1}{3}\) 

    \(\frac{4}{7} - \frac{1}{3} = x\)

   \( x = \frac{12}{21} - \frac{7}{21} = \frac{5}{21}\)