Question

1) 2x^3 - 8x = 0 2)2x (x - 15) - 4 (x - 15) = 0

Answer 1

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 \(x\in\left\{2;15\right\}\)

Answer 2

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 

\(2x\left(x-15\right)-4\left(x-15\right)=0\)    

\(\left(2x-4\right)\left(x-15\right)=0\)   

\(\orbr{\begin{cases}2x-4=0\\x-15=0\end{cases}}\)    

\(\orbr{\begin{cases}2x=4\\x=0+15\end{cases}}\)   

\(\orbr{\begin{cases}x=2\\x=15\end{cases}}\)

Answer 3

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

Answer 4

                                               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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