\(12x^3=3x\)
\(\Leftrightarrow12x^3-3x=0\)
\(\Leftrightarrow3x\left(4x^2-1\right)=0\)
\(\Leftrightarrow3x\left(2x+1\right)\left(2x-1\right)=0\)
\(\Leftrightarrow x\in\left\{0;\pm\frac{1}{2}\right\}\)
<=> \(3x\left(4x^2-1\right)=0\)
<=> \(\orbr{\begin{cases}x=0\\4x^2=1\end{cases}\Leftrightarrow\orbr{\begin{cases}x=0\\x^2=\frac{1}{4}\end{cases}}}\)
<=> \(\orbr{\begin{cases}x=0\\x=\pm\frac{1}{2}\end{cases}}\)
Vay \(x\in\left\{-\frac{1}{2};0;\frac{1}{2}\right\}\)
Hoc tot
\(12x^3=3x\)
<=>\(12x^3-3x=0\)
<=>\(3x\left(4x^2-1\right)=0\)
<=>\(3x\left(2x-1\right)\left(2x+1\right)=0\)
<=>\(3x=0=>x=0\)
\(\orbr{\begin{cases}2x-1=0\\2x+1=0\end{cases}}\)
<=>\(\orbr{\begin{cases}2x=1\\2x=-1\end{cases}}\)
<=>\(\orbr{\begin{cases}x=\frac{1}{2}\\x=\frac{-1}{2}\end{cases}}\)
Vậy \(x\in(0;\frac{1}{2};\frac{-1}{2})\)
\(12.x^3=3.x\)
\(\Leftrightarrow4.x^3=x\)
\(\Leftrightarrow4.x^3-x=0\)
\(\Leftrightarrow x.\left(4.x^2-1\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x=0\\4.x^2-1=0\end{cases}}\)\(\Rightarrow x\hept{\begin{cases}=0\\=-\frac{1}{2}\\=\frac{1}{2}\end{cases}}\)
Vậy \(x=\left\{0;-\frac{1}{2};\frac{1}{2}\right\}\)
\(12x^3=3x=>3x.4x^2=3x\)
=> \(4x^2=1\)=> \(x^2=\frac{1}{4}\)
=> \(\orbr{\begin{cases}x=\frac{1}{2}\\x=-\frac{1}{2}\end{cases}}\)
dễ thấy x=0 thì \(12x^3=3x=0\)
Vậy \(x=\left\{\frac{1}{2};0;-\frac{1}{2}\right\}\)
Trả lời:
\(12x^3=3x\)
\(\Leftrightarrow4x^3=x\)
\(\Leftrightarrow4x^3-x=0\)
\(\Leftrightarrow x\left(4x^2-1\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x=0\\4x^2-1=0\end{cases}}\)\(\Leftrightarrow\orbr{\begin{cases}x=0\\4x^2=1\end{cases}}\)\(\Leftrightarrow\orbr{\begin{cases}x=0\\x^2=\frac{1}{4}\end{cases}}\)
\(\Leftrightarrow\)\(x=0\)hoặc\(\orbr{\begin{cases}x=\frac{1}{2}\\x=\frac{-1}{2}\end{cases}}\)
Vậy\(x\in\left\{0;\frac{1}{2};\frac{-1}{2}\right\}\)
Hok tốt!
Good girl
