Tìm x :
Câu a : \(x\left(4x^2-1\right)=0\)
\(\Leftrightarrow x\left(2x-1\right)\left(2x+1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\2x-1=0\\2x+1=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=0\\x=\dfrac{1}{2}\\x=-\dfrac{1}{2}\end{matrix}\right.\)
Vậy \(x=0\) ; \(x=\dfrac{1}{2}\) hoặc \(x=-\dfrac{1}{2}\)
Câu b : \(3\left(x-1\right)^2-3x\left(x-5\right)-2=0\)
\(\Leftrightarrow3x^2-6x+3-3x^2+15x-2=0\)
\(\Leftrightarrow9x+1=0\Rightarrow x=-\dfrac{1}{9}\)
a) x3 - 4x2 + 12x - 27
= x3-3x2-x2+3x+9x-27
= ( x3-3x2)-(x2-3x)+(9x-27)
=x2(x-3)-x(x-3)+9(x-3)
= (x-3)(x2-x+9)
bài tìm x
c) x3 - x2 - x + 1 = 0
=>x3+x2-2x2-2x+x+1=0
=>(x3+x2)-(2x2+2x)+(x+1)=0
=>x2(x+1)-2x(x+1)+(x+1)=0
=> (x+1)(x2-2x+1)=0
=>(x+1)(x-1)2=0
=> \(\left[{}\begin{matrix}x+1=0\\x-1=0\end{matrix}\right.\) =>\(\left[{}\begin{matrix}x=-1\\x=1\end{matrix}\right.\)
vậy x\(\in\left\{-1;1\right\}\)
