c) \(\left|2x-\dfrac{3}{7}\right|-\dfrac{1}{2}=\dfrac{3}{4}\)
\(\left|2x-\dfrac{3}{7}\right|=\dfrac{3}{4}+\dfrac{1}{2}\)
\(\left|2x-\dfrac{3}{7}\right|=\dfrac{5}{4}\)
\(\Rightarrow\left\{{}\begin{matrix}2x-\dfrac{3}{7}=\dfrac{5}{4}\\2x-\dfrac{3}{7}=\dfrac{-5}{4}\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}2x=\dfrac{47}{28}\\2x=\dfrac{-23}{28}\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=\dfrac{47}{56}\\x=\dfrac{-23}{56}\end{matrix}\right.\)
Vậy \(x=\dfrac{47}{56}\) hoặc \(x=\dfrac{-23}{56}\)
d) \(\dfrac{2x+1}{3}=\dfrac{x-5}{2}\)
\(\Rightarrow\left(2x+1\right).2=\left(x-5\right).3\)
\(2x.2+1.2=x.3-5.3\)
\(4x-3x=\left(-15\right)-2\)
\(x=-17\)
Vậy \(x=-17\)
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