a) \(3\dfrac{1}{4}=\dfrac{2}{3}:\left(\dfrac{-x}{2}\right)\Leftrightarrow\dfrac{13}{4}=\dfrac{2}{3}.\dfrac{-2}{x}\Leftrightarrow\dfrac{-2}{x}=\dfrac{39}{8}\Leftrightarrow x=-\dfrac{16}{39}\)
b) \(1-2\left(x+\dfrac{1}{3}\right)=\left|-\dfrac{2}{3}+\dfrac{1}{5}\right|\Leftrightarrow1-2x-\dfrac{2}{3}=\dfrac{7}{15}\Leftrightarrow2x=-\dfrac{2}{15}\Leftrightarrow x=-\dfrac{1}{15}\)
c) \(\left(2x-1\right)\left(\dfrac{2}{5}-\dfrac{1}{3}x\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}2x-1=0\\\dfrac{2}{5}-\dfrac{1}{3}x=0\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1}{2}\\\dfrac{1}{3}x=\dfrac{2}{5}\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1}{2}\\x=\dfrac{6}{5}\end{matrix}\right.\)
d) \(-4\dfrac{3}{5}.2\dfrac{4}{23}\le x\le-2\dfrac{3}{5}:1\dfrac{6}{15}\Leftrightarrow-10\le x\le-\dfrac{13}{7}\Leftrightarrow x\in\left\{-10;-9;-8;-7;-6;-5;-4;-3;-2;-1\right\}\)(do \(x\in Z\))
Bài 2:
c: Ta có: \(\left(2x-1\right)\left(\dfrac{2}{5}-\dfrac{1}{3}x\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}2x-1=0\\\dfrac{2}{5}-\dfrac{1}{3}x=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}2x=1\\\dfrac{1}{3}x=\dfrac{2}{5}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1}{2}\\x=\dfrac{6}{5}\end{matrix}\right.\)
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