a, \(x\left(4x^2-1\right)=0\Leftrightarrow x\left[\left(2x\right)^2-1^2\right]=0\Leftrightarrow x\left(2x+1\right)\left(2x-1\right)=0\Leftrightarrow\left\{{}\begin{matrix}x=0\\2x+1=0\\2x-1=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=0\\x=\dfrac{-1}{2}\\x=\dfrac{1}{2}\end{matrix}\right.\)
b,
a)\(x\cdot\left(4x^2\right)=0\)\(\Rightarrow\left\{{}\begin{matrix}x=0\\4x^2-1=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=0\\\left(2x-1\right)\left(2x+1\right)=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=0\\x=\dfrac{1}{2}\\x=\dfrac{-1}{2}\end{matrix}\right.\)
a) x(4x2 - 1) = 0
x(2x - 1)(2x + 1) = 0
\(\Rightarrow\left\{{}\begin{matrix}x=0\\2x-1=0\\2x+1=0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=0\\2x=1\\2x=-1\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=0\\x=\dfrac{1}{2}\\x=\dfrac{-1}{2}\end{matrix}\right.\)Vậy x = 0 hoặc x = \(\dfrac{1}{2}\)hoặc x = -\(\dfrac{1}{2}\)
b) 3(x - 1)2 - 3x(x - 5) - 2 = 0
3(x2 - 2x + 1) - (3x2 - 15x) - 2 = 0
3x2 - 6x + 3 - 3x2 + 15x - 2 = 0
9x + 1 = 0
\(\Rightarrow\) 9x = -1
\(\Rightarrow\) x = \(\dfrac{-1}{9}\)
Vậy x = \(\dfrac{-1}{9}\)
c) x3 - x2 - x + 1 = 0
(x3 - x2) - (x - 1) = 0
x2(x - 1) - (x - 1) = 0
(x - 1)(x2 - 1) = 0
(x - 1)(x - 1)(x + 1) = 0
(x - 1)2(x + 1)
\(\Leftrightarrow\left\{{}\begin{matrix}\left(x-1\right)^2=0\\x+1=0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x-1=0\\x=-1\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=1\\x=-1\end{matrix}\right.\)Vậy x = 1 hoặc x = -1
d) 2x2 - 5x - 7 = 0
\(\Rightarrow\)2x2 + 2x - 7x - 7 = 0
\(\Rightarrow\)(2x2 + 2x) - (7x + 7) = 0
\(\Rightarrow\) 2x(x + 1) - 7(x + 1) = 0
\(\Rightarrow\)(2x - 7)(x + 1) = 0
\(\Leftrightarrow\left\{{}\begin{matrix}2x-7=0\\x+1=0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}2x=7\\x=-1\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=\dfrac{7}{2}\\x=-1\end{matrix}\right.\)Vậy x =... hoặc x = ...
