`2x . ( x - 1/7) = 0`
`=>` \(\left[{}\begin{matrix}2x=0\\x-\dfrac{1}{7}=0\end{matrix}\right.\)
`=>` \(\left[{}\begin{matrix}x=0:2\\x=0+\dfrac{1}{7}\end{matrix}\right.\)
`=>`\(\left[{}\begin{matrix}x=0\\x=\dfrac{1}{7}\end{matrix}\right.\)
Vậy......
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`11/12 - (2/5 + x) = 2/3`
`2/5 + x = 11/12 - 2/3`
`2/5 + x =1/4`
`x = 1/4 - 2/5`
`x= (-3)/20`
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`#BaoL i nh`
\(2x\cdot\left(x-\dfrac{1}{7}\right)=0\) \(\Leftrightarrow\left[{}\begin{matrix}2x=0\\x-\dfrac{1}{7}=0\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=\dfrac{1}{7}\end{matrix}\right.\)
\(\dfrac{11}{12}-\left(\dfrac{2}{5}+x\right)=\dfrac{2}{3}\) \(\Leftrightarrow\dfrac{11}{12}-\dfrac{2}{5}-x=\dfrac{2}{3}\)\(\Leftrightarrow x=-\dfrac{3}{20}\)
\(2x.\left(x-\dfrac{1}{7}\right)=0\\ < =>\left[{}\begin{matrix}2x=0\\x-\dfrac{1}{7}=0\end{matrix}\right.< =>\left[{}\begin{matrix}x=0\\x=\dfrac{1}{7}\end{matrix}\right.\)
\(\dfrac{11}{12}-\left(\dfrac{2}{5}+x\right)=\dfrac{2}{3}\\ < =>\dfrac{2}{5}+x=\dfrac{11}{12}-\dfrac{2}{3}=\dfrac{1}{4}< =>x=\dfrac{1}{4}-\dfrac{2}{5}=\dfrac{-3}{20}\)
