Bài 9 trang 10 SGK Toán 7 tập 1

Đề bài

Tìm \(x\), biết:

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

b) \(x - \dfrac{2}{5} = \dfrac{5}{7}\)

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

d) \(\dfrac{4}{7} - x = \dfrac{1}{3}\)

Lời giải

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

     \(x = \dfrac{3}{4} - \dfrac{1}{3} \)

     \(x= \dfrac{9}{12} - \dfrac{4}{12} \)

     \(x= \dfrac{5}{12}\)

b)  \(x - \dfrac{2}{5} = \dfrac{5}{7}\)

     \(x = \dfrac{5}{7} + \dfrac{2}{5} \)

     \(x= \dfrac{25}{35} + \dfrac{14}{35}\)

     \(x= \dfrac{39}{35} = 1\dfrac{4}{35}\)

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

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

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

    \(x= \dfrac{4}{21}\)

d) \(\dfrac{4}{7} - x = \dfrac{1}{3}\) 

    \(\dfrac{4}{7} - \dfrac{1}{3} = x\)

    \( x = \dfrac{12}{21} - \dfrac{7}{21}\)

    \(x= \dfrac{5}{21}\)


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