\(4x\left(5-x\right)-8x\left(5-x\right)=0\)
\(\Rightarrow x\left(5-x\right).\left(4-8\right)=0\)
\(\Rightarrow x\left(5-x\right)=0\)
\(\Rightarrow\left\{{}\begin{matrix}x=0\\5-x=0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=0\\x=5\end{matrix}\right.\)
Vậy \(x\in\left\{0;5\right\}\)
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Ta có:4x(5-x)-8x(5-x)=0<=>4x(5-x)-2.4x(5-x)<=>4x(5-x)(1-2)=0
<=>-4x(5-x)=0
=>4x=0 hoặc 5-x=0=>x=0 hoặc x=5
Vậy S={0;5}
<=> 20x-4x2-40x+8x2=0
<=> 4x2-20x=0
<=> x(4x-20)=0
<=> \(\left\{{}\begin{matrix}x=0\\4x-20=0< =>4x=20< =>x=5\end{matrix}\right.\)
\(4x\left(5-x\right)-8x\left(5-x\right)=0\\ \Leftrightarrow20x-4x^2-40x+8x^2=0\\ \Leftrightarrow4x^2-20x=0\\ \Leftrightarrow4x\left(x-5\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}4x=0\\x-5=0\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=5\end{matrix}\right.\)
vậy phương trình có tập nghiệm là S={0;5}
\(4x\left(5-x\right)-8x\left(5-x\right)=0\Leftrightarrow\left(4x-8x\right)\left(5-x\right)=0\Leftrightarrow-4x\left(5-x\right)=0\Leftrightarrow\left[{}\begin{matrix}-4x=0\\5-x=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=5\end{matrix}\right.\)