a, 3x(x-2)=7(x-2)
\(\Leftrightarrow\)3x(x-2)-7(x-2)=0
\(\Leftrightarrow\left(x-2\right)\left(3x-7\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}3x-7=0\\x-2=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\frac{7}{3}\\x=2\end{matrix}\right.\)
Vậy phương trình có tập nghiệm S=\(\left\{\frac{7}{3};2\right\}\)
b, 2x2=2019x
\(\Leftrightarrow\)2x2-2019x=0
\(\Leftrightarrow x\left(2x-2019\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\2x-2019=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=\frac{2019}{2}\end{matrix}\right.\)
Vậy phương trình có tập nghiệm S=\(\left\{0;\frac{2019}{2}\right\}\)
a)\(3x\left(x-2\right)=7\left(x-2\right)\)
\(\rightarrow3x\left(x-2\right)-7\left(x-2\right)=0\)
\(\rightarrow\left(3x-7\right)\left(x-2\right)=0\)
\(\rightarrow\left[{}\begin{matrix}3x-7=0\\x-2=0\end{matrix}\right.\)
\(\Rightarrow x\in\left\{\frac{7}{3};2\right\}\)
Vậy \(x\in\left\{\frac{7}{3};2\right\}\)
b)\(2x^2=2019x\)
\(\rightarrow2x^2-2019x=0\)
\(\rightarrow x\left(2x-2019\right)=0\)
\(\rightarrow\left[{}\begin{matrix}x=0\\2x-2019=0\end{matrix}\right.\)
\(\Rightarrow x\in\left\{0;\frac{2019}{2}\right\}\)
Vậy \(x\in\left\{0;\frac{2019}{2}\right\}\)
a) 3x(x-2)=7(x-2)
\(\Leftrightarrow3x^2-6x-7x+14=0\)
\(\Leftrightarrow3x^2-13x+14=0\)
\(\Leftrightarrow3x^2-6x-7x+14=0\)
\(\Leftrightarrow3x\left(x-2\right)-7\left(x-2\right)=0\)
\(\Leftrightarrow\left(x-2\right)\left(3x-7\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-2=0\\3x-7=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=2\\x=\frac{7}{3}\end{matrix}\right.\)
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