Giải Bài 14 trang 13 sách bài tập toán 7 tập 1 – Cánh diều

a)

\(\begin{array}{l}\dfrac{3}{7} + x =  – 1,5\\{\rm{   }}\dfrac{3}{7} + x = \dfrac{{ – 3}}{2}\\{\rm{    }}x = \dfrac{{ – 3}}{2} – \dfrac{3}{7}\\{\rm{    }}x = \dfrac{{ – 21}}{{14}} – \dfrac{6}{{14}}\\{\rm{    }}x = \dfrac{{ – 27}}{{14}}\end{array}\)

Vậy \(x = \dfrac{{ – 27}}{{14}}\).                                   

b)

\(\begin{array}{l}3\dfrac{1}{5} – x = 1,6 + \dfrac{7}{{10}}\\{\rm{  }}\dfrac{{16}}{5} – x = \dfrac{{16}}{{10}} + \dfrac{7}{{10}}\\{\rm{  }}\dfrac{{16}}{5} – x = \dfrac{{23}}{{10}}\\{\rm{  }}x = \dfrac{{16}}{5} – \dfrac{{23}}{{10}}\\{\rm{  }}x = \dfrac{{32}}{{10}} – \dfrac{{23}}{{10}}\\{\rm{  }}x = \dfrac{9}{{10}}\end{array}\)

Vậy \(x = \dfrac{9}{{10}}\).

c)

\(\begin{array}{l}x.\dfrac{{14}}{3} = 2,5\\{\rm{  }}x.\dfrac{{14}}{3} = \dfrac{5}{2}\\{\rm{  }}x = \dfrac{5}{2}:\dfrac{{14}}{3}\\{\rm{  }}x = \dfrac{5}{2}.\dfrac{3}{{14}}\\{\rm{  }}x = \dfrac{{15}}{{28}}\end{array}\)

Vậy \(x = \dfrac{{15}}{{28}}\).                                            

d)

\(\begin{array}{l}x:\left( { – \dfrac{3}{5}} \right) = 1\dfrac{1}{4}\\{\rm{  }}x:\left( { – \dfrac{3}{5}} \right) = \dfrac{5}{4}\\{\rm{  }}x = \dfrac{5}{4}.\left( { – \dfrac{3}{5}} \right)\\{\rm{  }}x =  – \dfrac{3}{4}\end{array}\)

Vậy \(x =  – \dfrac{3}{4}\).