Đề bài
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] \[{1 \over 3} + {3 \over {35}} < {x \over {210}} < {4 \over 7} + {3 \over 7} + {1 \over 3}\].
Lời giải chi tiết
\[a]{{ - 8} \over {17}} + {5 \over {17}} = {{ - 3} \over {17}};{{ - 6} \over {17}} + {9 \over {17}} = {3 \over {17}}.\]
Do đó: \[{{ - 3} \over {17}} < {x \over {17}} < {3 \over {17}} \Rightarrow - 3 < x < 3.\]
Mà \[x \in Z\] nên \[x \in \left\{ { - 2; - 1;0;1;2} \right\}\]
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 \[x \in \left\{ {89;90;91;...;278;279} \right\}\]