a ) \(x^3-\frac{1}{4}x=0\)
\(\Leftrightarrow x.\left(x^2-\frac{1}{4}\right)=0\)
\(\Leftrightarrow x.\left(x^2-\frac{1}{2}^2\right)=0\)
\(\Leftrightarrow x.\left(x-\frac{1}{2}\right).\left(x+\frac{1}{2}\right)=0\)
\(\Leftrightarrow\left[\begin{array}{nghiempt}x=0\\x-\frac{1}{2}=0\\x+\frac{1}{2}=0\end{array}\right.\Leftrightarrow\left[\begin{array}{nghiempt}x=0\\x=\frac{1}{2}\\x=-\frac{1}{2}\end{array}\right.\)
Vậy \(x\in\left\{0;\frac{1}{2};-\frac{1}{2}\right\}\)
b ) \(\left(2x-1\right)^2-\left(x+3\right)^2=0\)
\(\Leftrightarrow\left[\left(2x-1\right)-\left(x+3\right)\right].\left[\left(2x-1\right)+\left(x+3\right)\right]\)
\(\Leftrightarrow\left[2x-1-x+3\right].\left[2x-1+x+3\right]\)
\(\Leftrightarrow\left(x-4\right)\left(3x+2\right)=0\)
\(\Leftrightarrow\left[\begin{array}{nghiempt}x-4=0\\3x+2=0\end{array}\right.\)
\(\Leftrightarrow\left[\begin{array}{nghiempt}x=4\\x=-\frac{2}{3}\end{array}\right.\)
Vậy \(x\in\left\{4;-\frac{2}{3}\right\}\)
