Tìm x
a) 6X2_12X=0
6x(x-2)=0
=> \(\left[{}\begin{matrix}6x=0\\x-2=0\end{matrix}\right.\) => \(\left[{}\begin{matrix}x=0\\x=2\end{matrix}\right.\)
Vậy x \(\in\) {0;2}
b) 92-(2x-1)2=0
(2x-1-9).(2x-1+9)=0
(2x-10).(2x+8)=0
=> \(\left[{}\begin{matrix}2x-10=0\\2x+8=0\end{matrix}\right.=>\left[{}\begin{matrix}2x=10=>x=5\\2x=-8=>x=-4\end{matrix}\right.\)
Vậy x \(\in\left\{5;-4\right\}\)
c)3x(x-2)+4x-8=0
3x(x-2) + 4(x-2)=0
(x-2)(3x+4)=0
=> \(\left[{}\begin{matrix}x-2=0\\3x+4=0\end{matrix}\right.=>\left[{}\begin{matrix}x=2\\3x=-4=>x=-\dfrac{4}{3}\end{matrix}\right.\)
d) 2x2+5x-12=0
2x2+8x-3x-12=0
=> (2x2+8x)-(3x+12)=0
=> 2x(x+4)-3(x+4)=0
=> (x+4)(2x-3)
=> \(\left[{}\begin{matrix}x+4=0\\2x-3=0\end{matrix}\right.=>\left[{}\begin{matrix}x=-4\\2x=3=>x=\dfrac{3}{2}\end{matrix}\right.\)
=> \(x\in\left\{-4;\dfrac{3}{2}\right\}\)
a: =>6x(x-2)=0
=>x=0 hoặc x=2
b: =>(2x-1-9)(2x-1+9)=0
=>(2x-10)(2x+8)=0
=>x=-4 hoặc x=4
c: =>(x-2)(3x+4)=0
=>x=-4/3 hoặc x=2
