Advertisements (Quảng cáo)
Điền dấu (<, >, =) thích hợp vào ô trống :
a) \({{ – 5} \over 8} + {3 \over { – 8}}… 1\) ;
b) \({{ – 13} \over {22}} + {{ – 5} \over {22}}…{{ – 8} \over {11}}\);
c) \({1 \over 6} + {{ – 3} \over 4}…{1 \over {14}} + {{ – 4} \over 7}\) .
\(a){{ – 5} \over 8} + {3 \over { – 8}} = {{ – 5} \over 8} + {{ – 3} \over 8} = {{ – 8} \over 8} = – 1.\)
Advertisements (Quảng cáo)
Vậy \({{ – 5} \over 8} + {3 \over { – 8}} = – 1\)
\(b){{ – 13} \over {22}} + {{ – 5} \over {22}} = {{ – 18} \over {22}} = {{ – 9} \over {11}} < {{ – 8} \over {11}}.\)
Vậy \({{ – 13} \over {22}} + {{ – 5} \over {22}} < {{ – 8} \over {11}}.\)
\(\eqalign{ & c){1 \over 6} + {{ – 3} \over 4} = {2 \over {12}} + {{ – 9} \over {12}} = {{ – 7} \over {12}} \cr & {1 \over {14}} + {{ – 4} \over 7} = {1 \over {14}} + {{ – 8} \over {14}} = {{ – 7} \over {14}} = {{ – 1} \over 2} = {{ – 6} \over {12}} \cr} \)
Vì \({{ – 6} \over {12}} > {{ – 7} \over {12}}\) nên \({1 \over 6} + {{ – 3} \over 4} < {1 \over {14}} + {{ – 4} \over 7}\)