a) \(\dfrac{1}{2}x-\dfrac{3}{4}x-\dfrac{7}{3}=-\dfrac{5}{6}\\ < =>-\dfrac{1}{4}x-\dfrac{7}{3}=-\dfrac{5}{6}\\ < =>-\dfrac{1}{4}x=-\dfrac{5}{6}+\dfrac{7}{3}=\dfrac{3}{2}\\ =>x=\dfrac{3}{2}:\dfrac{-1}{4}=-6\)
b) \(\left|x-\dfrac{1}{6}\right|+-\dfrac{5}{12}=\dfrac{4}{7}.\dfrac{14}{48}\\ < =>\left|x-\dfrac{1}{6}\right|=\left(\dfrac{4}{7}.\dfrac{14}{48}\right)-\left(-\dfrac{5}{12}\right)=\dfrac{1}{6}+\dfrac{5}{12}=\dfrac{7}{12}\\ \)
Xảy ra 2 trường hợp:
+) TH1: \(x-\dfrac{1}{6}=\dfrac{7}{12}\\ =>x=\dfrac{7}{12}+\dfrac{1}{6}=\dfrac{3}{4}->\left(a\right)\)
+) TH2" \(-\left(x-\dfrac{1}{6}\right)=\dfrac{7}{12}\\ < =>-x+\dfrac{1}{6}=\dfrac{7}{12}\\ < =>-x=\dfrac{7}{12}-\dfrac{1}{6}=\dfrac{5}{12}\\ =>x=-\dfrac{5}{12}->\left(b\right)\)
Từ (a) và (b) => \(x\in\left\{-\dfrac{5}{12};\dfrac{3}{4}\right\}\)
a, \(\dfrac{1}{2}x-\dfrac{3}{4}x-\dfrac{7}{3}=\dfrac{-5}{6}\)
\(\Rightarrow\dfrac{-1}{4}x=\dfrac{-5}{6}+\dfrac{7}{3}\)
\(\Rightarrow\dfrac{-1}{4}x=\dfrac{3}{2}\Rightarrow x=-6\)
Vậy \(x=-6\)
b, \(\left|x-\dfrac{1}{6}\right|+\dfrac{-5}{12}=\dfrac{4}{7}.\dfrac{14}{48}\)
\(\Rightarrow\left|x-\dfrac{1}{6}\right|-\dfrac{5}{12}=\dfrac{1}{6}\)
\(\Rightarrow\left|x-\dfrac{1}{6}\right|=\dfrac{7}{12}\)
\(\Rightarrow\left\{{}\begin{matrix}x-\dfrac{1}{6}=\dfrac{-7}{12}\\x-\dfrac{1}{6}=\dfrac{7}{12}\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=\dfrac{-5}{12}\\x=\dfrac{3}{4}\end{matrix}\right.\)
Vậy \(x\in\left\{\dfrac{-5}{12};\dfrac{3}{4}\right\}\)
Chúc bạn học tốt!!!
\(a,\dfrac{1}{2}x-\dfrac{3}{4}x-\dfrac{7}{3}=-\dfrac{5}{6}\\ \dfrac{1}{2}x-\dfrac{3}{4}x=\dfrac{-5}{6}+\dfrac{7}{3}\\ x\left(\dfrac{1}{2}-\dfrac{3}{4}\right)=\dfrac{-5}{6}+\dfrac{14}{6}\\ x\left(\dfrac{2}{4}-\dfrac{3}{4}\right)=\dfrac{9}{6}\\ \dfrac{-1}{4}x=\dfrac{3}{2}\\ x=\dfrac{3}{2}:\dfrac{-1}{4}\\ x=\dfrac{3}{2}\cdot\left(-4\right)\\ x=-6\)
a)\(\dfrac{1}{2}x-\dfrac{3}{4}x-\dfrac{7}{3}=\dfrac{-5}{6}\)
\(\dfrac{1}{2}x-\dfrac{3}{4}x=\dfrac{-5}{6}+\dfrac{7}{3}\)
\(\dfrac{1}{2}x-\dfrac{3}{4}x=\dfrac{3}{2}\)
\(x\left(\dfrac{1}{2}-\dfrac{3}{4}\right)=\dfrac{3}{2}\)
\(x.\dfrac{-1}{4}=\dfrac{3}{2}\)
\(x=\dfrac{3}{2}:\dfrac{-1}{4}\)
\(x=-6\)
\(\left|x-\dfrac{1}{6}\right|\)\(+\dfrac{-5}{12}=\)\(\dfrac{4}{7}.\dfrac{14}{48}\)
\(\left|x-\dfrac{1}{6}\right|+\dfrac{-5}{12}\)\(=\)\(\dfrac{1}{6}\)
\(\left|x-\dfrac{1}{6}\right|=\dfrac{1}{6}-\dfrac{-5}{12}\)
\(\left|x-\dfrac{1}{6}\right|=\dfrac{7}{12}\)
\(\Leftrightarrow x-\dfrac{1}{6}=\dfrac{7}{12}\Rightarrow x=\dfrac{7}{12}+\dfrac{1}{6}=\dfrac{3}{4}\)
\(x-\dfrac{1}{6}=\dfrac{-7}{12}\Rightarrow x=\dfrac{-7}{12}+\dfrac{1}{6}=\dfrac{-5}{12}\)
Vậy...
