\(8x.\left(5-x\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}8x=0\\5-x=0\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}x=0\\x=5\end{cases}}\)
Vậy \(x\in\left\{0;5\right\}\)
Có: 8x(5-x)=0
TH1:8x=0. TH2: 5-x =0
=>x=0. =>x=5
Vậy x=0 hoặc x=5
\(8x\cdot\left(5-x\right)=0\)
\(\)Có 2 trường hợp xảy ra :
\(TH1:8x=0\)
\(\Rightarrow\)\(x=0:8\)
\(x=0\)
\(TH2:5-x=0\)
\(x=0+5\)
\(x=5\)
Vậy \(x\in\left\{0;5\right\}\)
\(8x\left(5-x\right)=0\)
\(\Rightarrow\orbr{\begin{cases}8x=0\\5-x=0\end{cases}\Rightarrow\orbr{\begin{cases}x=0\\x=5\end{cases}\Rightarrow}x=0;5}\)\
Vậy x=0;5 nha
nhớ k mk đó@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@
