a) \(\left|1-x\right|=\dfrac{1}{4}\)
\(\Leftrightarrow\left[{}\begin{matrix}1-x=\dfrac{1}{4}\\1-x=-\dfrac{1}{4}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}-x=\dfrac{1}{4}-1\\-x=-\dfrac{1}{4}-1\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}-x=-\dfrac{3}{4}\\-x=-\dfrac{5}{4}\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{3}{4}\\x=\dfrac{5}{4}\end{matrix}\right.\)
Vậy: \(x=\dfrac{3}{4};\dfrac{5}{4}\)
b) \(\dfrac{x+1}{3}=\dfrac{x}{2}\)
\(\Leftrightarrow2\left(x+1\right)=3x\)
\(\Leftrightarrow2x+2=3x\)
\(\Leftrightarrow2x-3x=-2\)
\(\Leftrightarrow-x=-2\)
\(\Leftrightarrow x=2\)
Vậy: \(x=2\)
c) \(\dfrac{2x}{3}-\dfrac{x}{4}=\dfrac{5}{6}\)
\(\Leftrightarrow8x-3x=10\)
\(\Leftrightarrow5x=10\)
\(\Leftrightarrow x=2\)
Vậy: \(x=2\)
a) Ta có: \(\left|1-x\right|=\dfrac{1}{4}\)
\(\Rightarrow\left[{}\begin{matrix}1-x=\dfrac{1}{4}\\1-x=-\dfrac{1}{4}\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}x=1-\dfrac{1}{4}\\x=1-\left(-\dfrac{1}{4}\right)\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=\dfrac{3}{4}\\x=\dfrac{5}{4}\end{matrix}\right.\)
b) Vì \(\dfrac{x+1}{3}=\dfrac{x}{2}\Leftrightarrow2\left(x+1\right)=3x\)
\(\Leftrightarrow2x+2=3x\)
\(\Leftrightarrow x=3x-2x\)
\(\Leftrightarrow x=2\)
c) Ta có: \(\dfrac{2x}{3}-\dfrac{x}{4}=\dfrac{5}{6}\)
\(\Rightarrow\dfrac{8x}{12}-\dfrac{3x}{12}=\dfrac{10}{12}\)
\(\Rightarrow\dfrac{8x-3x}{12}=\dfrac{10}{12}\)
\(\Rightarrow\dfrac{x\left(8-3\right)}{12}=\dfrac{10}{12}\)
\(\Rightarrow\dfrac{5x}{12}=\dfrac{10}{12}\Rightarrow5x=10\Rightarrow x=2\)
Vậy \(x=2\)
