Advertisements (Quảng cáo)
a) \({{ – 8} \over {17}} + {5 \over {17}} < {x \over {17}} < {{ – 6} \over {17}} + {9 \over {17}}\);
b) \({{ – 8} \over {15}} < {x \over {40}} < {{ – 7} \over {15}}\);
c) \({1 \over 3} + {3 \over {35}} < {x \over {210}} < {4 \over 7} + {3 \over 7} + {1 \over 3}\).
\(a){{ – 8} \over {17}} + {5 \over {17}} = {{ – 3} \over {17}};{{ – 6} \over {17}} + {9 \over {17}} = {3 \over {17}}.\)
Do đó: \({{ – 3} \over {17}} < {x \over {17}} < {3 \over {17}} \Rightarrow – 3 < x < 3.\)
Mà \(x \in Z\) nên \(x \in \left\{ { – 2; – 1;0;1;2} \right\}\)
Advertisements (Quảng cáo)
b)Ta có: \({{ – 8} \over {15}} = {{ – 64} \over {120}};{x \over {40}} = {{3x} \over {120}};{{ – 7} \over {15}} = {{ – 56} \over {120}}.\)
Do đó: \({{ – 64} \over {120}} < {{3x} \over {120}} < {{ – 56} \over {120}} \Rightarrow – 64 < 3x < – 56.\)
Mà \(x \in Z\) nên \(3x \vdots 3.\) Do đó: \(3x \in \left\{ { – 63; – 60; – 57} \right\} \Rightarrow x \in \left\{ { – 21; – 20; – 19} \right\}\)
\(\eqalign{ & c){1 \over 3} + {3 \over {35}} = {{70} \over {210}} + {{18} \over {210}} = {{88} \over {210}} \cr & {4 \over 7} + {3 \over 7} + {1 \over 3} = {{120} \over {210}} + {{90} \over {210}} + {{70} \over {210}} = {{280} \over {210}} \cr} \)
Do đó: \({{88} \over {210}} < {x \over {210}} < {{280} \over {210}} \Rightarrow 88 < x < 280\)
Mà \(x \in Z\) nên \(x \in \left\{ {89;90;91;…;278;279} \right\}\)