Advertisements (Quảng cáo)
a) \({3 \over 7} + … = {{ – 2} \over 7}\) ;
b) \(… + {{ – 4} \over {11}} = {{ – 7} \over {11}}\) ;
c) \({{ – 6} \over {18}} + … = {{ – 3} \over {18}}\) ;
d) \({{ – 5} \over {13}} + … = {{ – 5} \over {13}}\).
Advertisements (Quảng cáo)
\(\eqalign{ & a){3 \over 7} + … = {2 \over 7} \cr & … = {{ – 2} \over 7} – {3 \over 7} = {{ – 5} \over 7} \cr} \)
Vậy \({3 \over 7} + {{ – 5} \over 7} = {{ – 2} \over 7}\)
\(\eqalign{ & b)… + {{ – 4} \over {11}} = {{ – 7} \over {11}} \cr & … = {{ – 7} \over {11}} + {4 \over {11}} = {{ – 3} \over {11}} \cr & \Rightarrow {{ – 3} \over {11}} + {{ – 4} \over {11}} = {{ – 7} \over {11}} \cr & c){{ – 6} \over {18}} + … = {{ – 3} \over {18}} \cr & … = {{ – 3} \over {18}} + {6 \over {18}} = {3 \over {18}} = {1 \over 6} \cr & \Rightarrow {{ – 6} \over {18}} + {1 \over 6} = {{ – 3} \over {18}}. \cr & d){{ – 5} \over {13}} + … = {{ – 5} \over {13}} \cr & … = {{ – 5} \over {13}} + {5 \over {13}} = 0 \cr & \Rightarrow {{ – 5} \over {13}} + 0 = {{ – 5} \over {13}}. \cr} \)