\(\left|2x-7\right|+\frac{1}{3}=1\frac{1}{2}\\ \left|2x-7\right|=1\frac{1}{2}-\frac{1}{3}\\ \left|2x-7\right|=\frac{7}{6}\)
Ta luôn biết: \(\forall x\in Q\) thì \(\left|x\right|=\left|-x\right|\)
\(\Rightarrow\left[{}\begin{matrix}2x-7=\frac{7}{6}\\-\left(2x-7\right)=\frac{7}{6}\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}2x=\frac{7}{6}+7\\2x-7=-\frac{7}{6}\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}2x=\frac{49}{6}\\2x=-\frac{7}{6}+7\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=\frac{49}{6}:2\\2x=\frac{35}{6}\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=\frac{49}{12}\\x=\frac{35}{6}:2\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=\frac{49}{12}\\x=\frac{35}{12}\end{matrix}\right.\)
Vậy \(x\in\left\{\frac{49}{12};\frac{35}{12}\right\}\)
k tính sai rồi mong bạn thông cảm nhé, làn sau mk cẩn thận hơn
\(\left|2x-7\right|+\frac{1}{3}=1\frac{1}{2}\)
\(\left|2x-7\right|+\frac{1}{3}=\frac{3}{2}\)
\(\left|2x-7\right|=\frac{3}{2}-\frac{1}{3}\)
\(\left|2x-7\right|=\frac{7}{6}\)
=> 2x - 7 = \(\frac{7}{6}\) hoặc 2x - 7 = \(\frac{-7}{6}\)
2x = \(\frac{49}{6}\) 2x = \(\frac{35}{6}\)
x = \(\frac{49}{12}\) x = \(\frac{35}{12}\)
\(\left|2x-7\right|+\frac{1}{3}=1\frac{1}{2}\)
\(\Leftrightarrow\left|2x-7\right|=\frac{11}{6}\)
*Th1 : 2x-7=\(\frac{11}{6}\)khi 2x-7\(\ge0\)
\(\Leftrightarrow\)2x=\(\frac{53}{6}\) khi 2x\(\ge7\)
\(\Leftrightarrow\)x= \(\frac{53}{12}\) khi x\(\ge\frac{7}{2}\)(tm)
*th2 : -2x+7=\(\frac{11}{6}\) khi 2x-7\(\le0\)
\(\Leftrightarrow\)-2x=\(-\frac{31}{6}\) khi 2x\(\le7\)
\(\Leftrightarrow\)x=\(\frac{31}{12}\) khi x\(\le\frac{7}{2}\)(ko tm)
Vậy x=\(\frac{53}{12}\)
