Tìm x:
a) (5x - 1)2 - (x2 - 4x + 4) = 0
\(\Leftrightarrow\left(5x-1\right)^2-\left(x-2\right)^2=0\)
\(\Leftrightarrow\left(5x-1-x+2\right)\left(5x-1+x-2\right)=0\)
\(\Leftrightarrow\left(4x+1\right)\left(6x-3\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}4x+1=0\\6x-3=0\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}4x=-1\\6x=3\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x=-\frac{1}{4}\\x=\frac{1}{2}\end{matrix}\right.\)
Vậy \(x=\left\{-\frac{1}{4};\frac{1}{2}\right\}\)
b) (4x - 1).(x + 3) = x2 - 9
\(\Leftrightarrow\left(4x-1\right)\left(x+3\right)-\left(x-3\right)\left(x+3\right)=0\)
\(\Leftrightarrow\left(x+3\right)\left(4x-1-x+3\right)=0\)
\(\Leftrightarrow\left(x+3\right)\left(3x+2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x+3=0\\3x+2=0\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x=-3\\x=-\frac{2}{3}\end{matrix}\right.\)
Vậy \(x=\left\{-3;-\frac{2}{3}\right\}\)
c) x3 - 3x + 2 = 0
\(\Leftrightarrow x^3-x-2x+2=0\)
\(\Leftrightarrow x\left(x^2-1\right)-2\left(x-1\right)=0\)
\(\Leftrightarrow x\left(x-1\right)\left(x+1\right)-2\left(x-1\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left[x\left(x+1\right)-2\right]=0\)
\(\Leftrightarrow\left(x-1\right)\left(x^2+x-2\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left(x^2-x+2x-2\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left[x\left(x-1\right)+2\left(x-1\right)\right]=0\)
\(\Leftrightarrow\left(x-1\right)^2\left(x+2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-1=0\\x+2=0\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x=1\\x=-2\end{matrix}\right.\)
Vậy x ={1; -2}
