Advertisements (Quảng cáo)
Tìm x, biết :
a) \(x = {{ – 1} \over 3} + {5 \over 6}\) ;
b) \(x + {4 \over 5} = {3 \over {10}}\) ;
c) \(x – {3 \over 4} = {1 \over 2}\) ;
d) \({{ – 5} \over 6} – x = {7 \over {12}} + {{ – 1} \over 2}\) ;
Advertisements (Quảng cáo)
e) \({4 \over 7} – \left( { – x} \right) = {{ – 1} \over {14}}\).
\(a)x = {{ – 1} \over 3} + {5 \over 6} = {{ – 2} \over 6} + {5 \over 6} = {3 \over 6} = {1 \over 2}.\) Vậy \(x = {1 \over 2}.\)
\(\eqalign{ & b)x + {4 \over 5} = {3 \over {10}} \Rightarrow x = {3 \over {10}} – {4 \over 5} = {3 \over {10}} – {8 \over {10}} = {{ – 5} \over {10}} = {{ – 1} \over 2}. \cr & c)x – {3 \over 4} = {1 \over 2} \Rightarrow x = {1 \over 2} + {3 \over 4} = {2 \over 4} + {3 \over 4} = {5 \over 4}. \cr & d){{ – 5} \over 6} – x = {7 \over {12}} + {{ – 1} \over 2} \Rightarrow – x = {7 \over {12}} + {{ – 1} \over 2} + {5 \over 6} = {7 \over {12}} + {{ – 6} \over {12}} + {{10} \over {12}} = {{7 – 6 + 10} \over {12}} = {{11} \over {12}} \Rightarrow x = {{ – 11} \over {12}}. \cr & e){4 \over 7} – ( – x) = {{ – 1} \over {14}} \Rightarrow – ( – x) = {{ – 1} \over {14}} – {4 \over 7} = {{ – 1} \over {14}} – {8 \over {14}} = {{ – 1 – 8} \over {14}} = {{ – 9} \over {14}} \Rightarrow x = {{ – 9} \over {14}}. \cr} \)