Question

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Answer 1

Ta có \(\left|2x-1\right|=\dfrac{3}{2}\)

\(\Leftrightarrow\left[{}\begin{matrix}2x-1=\dfrac{3}{2}\\2x-1=-\dfrac{3}{2}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{5}{4}\\x=-\dfrac{1}{4}\end{matrix}\right.\)

=> \(\left[{}\begin{matrix}C=4.\dfrac{5}{4}+3\\C=4.\dfrac{-1}{4}+3\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}C=8\\C=2\end{matrix}\right.\)

b) Ta có C = \(-\dfrac{5}{2}\)

\(\Rightarrow4x+3=-\dfrac{5}{2}\Leftrightarrow x=-\dfrac{11}{8}\)

Answer 2

b) Để \(C=-\dfrac{5}{2}\) thì \(4x+3=\dfrac{-5}{2}\)

\(\Leftrightarrow4x=\dfrac{-5}{2}-3=\dfrac{-5}{2}-\dfrac{6}{2}=-\dfrac{11}{2}\)

hay \(x=-\dfrac{11}{8}\)