\(\left(\dfrac{1}{3}x+\dfrac{4}{3}\right)\left(-\dfrac{1}{5}x+\dfrac{3}{5}\right)=0\)
⇔ \(\dfrac{1}{3}x+\dfrac{4}{3}=0\) hoặc \(-\dfrac{1}{5}x+\dfrac{3}{5}=0\)
⇔ \(x=-4\) hoặc \(x=3\)
Vậy x = - 4 hoặ x = 3
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\(\left(\dfrac{1}{3}x+\dfrac{4}{3}\right)\left(-\dfrac{1}{5}x+\dfrac{3}{5}\right)=0\\ \Rightarrow\left[{}\begin{matrix}\dfrac{1}{3}x+\dfrac{4}{3}=0\\-\dfrac{1}{5}x+\dfrac{3}{5}=0\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=-4\\x=3\end{matrix}\right.\)
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