Ta có: \(\left|x-2.3\right|=\left|3x+1.3\right|\)
\(\Leftrightarrow\left[{}\begin{matrix}x-2.3=3x+1.3\\x-2.3=-3x-1.3\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}-2x=3.6\\4x=1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-1.8\\x=\dfrac{1}{4}\end{matrix}\right.\)
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`|x-2,3|=|3x+1,3|`
TH1
`x-2,3=3x+1,3`
`<=>-2x=3,6`
`<=>x=-1,8`
TH2
`-(x-2,3)=3x+1,3`
`<=>-x+2,3=3x+1,3`
`<=>-4x=-1`
`<=>x=1/4`
Vậy `x \in {-1,8 ; 1/4}`
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