a/ \(x\left(x-2\right)+x-2=0\)
\(\Leftrightarrow\left(x-2\right)\left(x+1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-2=0\\x+1=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=2\\x=-1\end{matrix}\right.\)
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b/ \(5x\left(x-3\right)-x+3=0\)
\(\Leftrightarrow\left(5x-1\right)\left(x-3\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}5x-1=0\\x-3=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1}{5}\\x=3\end{matrix}\right.\)
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a. x.(x - 2) + x - 2 = 0
\(\Leftrightarrow\)x(x-2)+(x-2)=0
\(^{_{ }\Leftrightarrow}\)(x-2)(x+1)=0
\(\left[{}\begin{matrix}x-2=0\\x+1=0\end{matrix}\right.\Leftrightarrow\)\(\left[{}\begin{matrix}x=2\\x=-1\end{matrix}\right.\)
Vậy x\(\in\)\(\left\{2;-1\right\}\)
b. 5x(x-3)-(x+3)
\(^{_{ }\Leftrightarrow}\)5x(x-3) + (x-3) = 0
\(^{_{ }\Leftrightarrow}\)(x-3)(5x+1) = 0
\(\Rightarrow\)\(\left\{{}\begin{matrix}x-3=0\\5x+1=0\end{matrix}\right.\)\(\Rightarrow\)\(\left\{{}\begin{matrix}x=3\\x=\dfrac{-1}{5}\end{matrix}\right.\)
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