Bài 1:
a) \({{17} \over {16}} - \left( {x - {7 \over 6}} \right) = {7 \over 4} \)
\(\;\;\Rightarrow {{17} \over 6} - x + {7 \over 6} = {7 \over 4}\)
\(\eqalign{& \Rightarrow - x = {7 \over 4} - {{17} \over 6} - {7 \over 6} \cr & \Rightarrow - x = {{42 - 68 - 28} \over {24}} \cr & \Rightarrow - x = {{ - 54} \over {24}} \cr& \Rightarrow x = {{54} \over {24}} = {9 \over 4}. \cr} \)
b) \({3 \over {35}} - \left( {{3 \over 5} - x} \right) = {2 \over 7} \)
\(\;\;\Rightarrow {3 \over {35}} - {3 \over 5} + x = {2 \over 7}\)
\(\eqalign{ & \Rightarrow x = {2 \over 7} - {3 \over {35}} + {3 \over 5} \cr & \Rightarrow x = {{28} \over {35}} = {4 \over 5}. \cr} \)
Bài 2: Ta có: \({3 \over 4} - {5 \over 6} = {{9 - 10} \over {12}} = {{ - 1} \over {12}};\)
\(1 - \left( {{2 \over 3} - {1 \over 4}} \right) = 1 - {2 \over 3} + {1 \over 4} = {{12 - 8 + 3} \over {12}}\)\(\; = {7 \over {12}}.\)
Vậy \({{ - 1} \over {12}} \le x < {7 \over {12}}.\)
Vì \(x \in\mathbb Z\) nên \(x \in \left\{{ - 1;0;1;2;3;4;5;6} \right\}.\)
