a, (x-2016)(2x-1)=0
<=>x=2016 hoặc x=-1/2
b, (x+2)(x+2-x+2)=0
<=>4(x+2)=0
<=>x+2=0
<=>x=-2
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a) 2x(x-2016)-x+2016=0
=>2x(x-2016)-(x-2016)=0
=>(x-2016)(2x-1)=0
=>\(\left\{{}\begin{matrix}x-2016=0\\2x-1=0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=2016\\x=\dfrac{1}{2}\end{matrix}\right.\)
vậy x=2016 hoặc x=\(\dfrac{1}{2}\)
b) (x+2)2-(x-2)(x+2)=0
=>(x+2)[(x+2)-(x-2)]=0
=>(x+2)(x+2-x+2)=0
=>(x+2)4=0
=>x+2=0
=>x=-2
vậy x=-2
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