Bài 1:
a) \( x:\left( {{2 \over 9} - {1 \over 5}} \right) = {8 \over {16}}\)
\(\Rightarrow x :{1 \over {45}} = {8 \over {16}}\)
\(\Rightarrow x = {8 \over {16}}.{1 \over {45}}\)
\(\Rightarrow x = {1 \over {90}}.\)
b) \({1 \over 3} + {1 \over 2}:x = {1 \over 5} \)
\(\Rightarrow {1 \over 2}:x = {1 \over 5} - {1 \over 3}\)
\( \Rightarrow {1 \over 2}:x = - {2 \over {15}}\)
\(\Rightarrow x = {1 \over 2}:\left( {{{ - 2} \over {15}}} \right) \)
\(\Rightarrow x = {1 \over 2}.\left( {{{ - 15} \over 2}} \right)\)
\(\Rightarrow x = {{ - 15} \over 4}. \)
Bài 2:
\(2{3 \over 4}:2{1 \over {16}}.\left( {{{ - 3} \over 5}} \right) + 4,5.{4 \over 5} \)
\(= {{11} \over 4}:{{33} \over {16}}.\left( {{{ - 3} \over 5}} \right) + {{45} \over {10}}.{4 \over 5}\)
\( = {{11} \over 4}.{{16} \over {33}}\left( { - {3 \over 5}} \right) + {9 \over 2}.{4 \over 5} \)
\(= - {4 \over 5} + {{18} \over 5} = {{ - 4 + 18} \over 5} = {{14} \over 5}.\)
Bài 3: Ta thấy: \({{x + 3} \over {x - 5}} < 0\) khi \(x + 3\) và \(x - 5\) trái dấu:
Trường hợp 1:
\(\left\{ \matrix {x + 3 > 0 \hfill \cr x - 5 < 0 \hfill \cr} \right. \Rightarrow \left\{ \matrix{x > - 3 \hfill \cr x < 5 \hfill \cr} \right.\)\(\; \Rightarrow - 3 < x < 5\)
Trường hợp 2:
\(\left\{ \matrix{ x + 3 < 0 \hfill \cr x - 5 > 0 \hfill \cr} \right. \Rightarrow \left\{ \matrix{x < - 3 \hfill \cr x > 5 \hfill \cr} \right. \Rightarrow x \in \emptyset \) ‘
Vậy: \({{x + 3} \over {x - 5}} < 0\) khi \( - 3 < x < 5.\)
