Question

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Answer 1

Ta có: \(\left(9x-4\right)\left(x-\dfrac{1}{2}\right)-\left(x-\dfrac{1}{2}\right)\left(6+x\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}9x-4-6-x=0\\x-\dfrac{1}{2}=0\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}8x=10\\x=\dfrac{1}{2}\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{5}{4}\\x=\dfrac{1}{2}\end{matrix}\right.\)

Answer 2

\(x\in\left\{\dfrac{5}{4};\dfrac{1}{2}\right\}\)

Answer 3

\(\Leftrightarrow\left(x-\dfrac{1}{2}\right).\left(9x-4-\left(6+x\right)\right)=0\)

\(\Leftrightarrow\left(x-\dfrac{1}{2}\right)\left(9x-4-6-x\right)=0\)

\(\Leftrightarrow\left(x-\dfrac{1}{2}\right)\left(8x-10\right)=0\\ \Leftrightarrow\left\{{}\begin{matrix}x-\dfrac{1}{2}=0\\8x-10=0\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}x=\dfrac{1}{2}\\x=\dfrac{10}{8}=\dfrac{5}{4}\end{matrix}\right.\)