1) \(2x^3-8x=0\)
\(\Leftrightarrow2x\left(x^2-4\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x=0\\x^2-4=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=0\\x^2=4\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=0\\x=\pm2\end{cases}}\)
Vậy \(x\in\left\{0;\pm2\right\}\)
2) \(2x\left(x-15\right)-4\left(x-15\right)=0\)
\(\Leftrightarrow\left(2x-4\right)\left(x-15\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}2x-4=0\\x-15=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=2\\x=15\end{cases}}\)
Vậy \(x\in\left\{2;15\right\}\)
1
\(2x^3-8x=0\)
\(2x\left(x^2-4\right)=0\)
\(\orbr{\begin{cases}2x=0\\x^2-4=0\end{cases}}\)
\(\orbr{\begin{cases}x=0\\x^2=4\end{cases}}\)
\(\orbr{\begin{cases}x=0\\x=\pm2\end{cases}}\)
2
\(2x\left(x-15\right)-4\left(x-15\right)=0\)
\(\left(2x-4\right)\left(x-15\right)=0\)
\(\orbr{\begin{cases}2x-4=0\\x-15=0\end{cases}}\)
\(\orbr{\begin{cases}2x=4\\x=0+15\end{cases}}\)
\(\orbr{\begin{cases}x=2\\x=15\end{cases}}\)
1) 2x3 - 8x = 0
<=> 2x( x2 - 4 ) = 0
<=> 2x( x - 2 )( x + 2 ) = 0
<=> 2x = 0 hoặc x - 2 = 0 hoặc x + 2 = 0
<=> x = 0 hoặc x = ±2
2) 2x( x - 15 ) - 4( x - 15 ) = 0
<=> ( x - 15 )( 2x - 4 ) = 0
<=> x - 15 = 0 hoặc 2x - 4 = 0
<=> x = 15 hoặc x = 2
Bài giải
1) \(2x^3-8x=0\)
\(2x\left(x^2-4\right)=0\)
\(\Rightarrow\orbr{\begin{cases}2x=0\\x^2-4=0\end{cases}}\Rightarrow\orbr{\begin{cases}x=0\\x=\pm2\end{cases}}\)
\(\Rightarrow\text{ }x\in\left\{0\text{ ; }\pm2\right\}\)
2) \(2x\left(x-15\right)-4\left(x-15\right)=0\)
\(2\left(x-15\right)\left(x-2\right)=0\)
\(\Rightarrow\text{ }x\in\left\{15\text{ ; }2\right\}\)
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