a, \(6⋮\left(x-1\right)\\ =>\left(x-1\right)\inƯ\left(6\right)=\left\{1;2;3;6\right\}\\ =>\left[{}\begin{matrix}x-1=1\\x-1=2\\x-1=3\\x-1=6\end{matrix}\right.=>\left[{}\begin{matrix}x=2\\x=3\\x=4\\x=7\end{matrix}\right.\\ =>x\in\left\{2;3;4;7\right\}\)
b, \(14⋮\left(2x+3\right)\\ =>\left(2x+3\right)\inƯ\left(14\right)=\left\{1;2;7;14\right\}\\ =>\left[{}\begin{matrix}2x+3=1\\2x+3=2\\2x+3=7\\2x+3=14\end{matrix}\right.=>\left[{}\begin{matrix}x=-1\left(loại\right)\\x=-\dfrac{1}{2}\left(loại\right)\\x=2\left(nhận\right)\\x=\dfrac{11}{2}\left(loại\right)\end{matrix}\right.\\ =>x=2\)
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