Câu hỏi của Yến Chử

Question

tìm x biết$\left(x+3\right)^{2014}=\left(x+3\right)^{2012}$

TV Cuber · Accepted Answer

`(x+3)^2014 = (x+3)^2012``(x+3)^2014 -(x+3)^2012 =0``(x+3)^2012 [(x+3)^2 -1]=0`TH1 :`(x+3)^2012 =0 => x+3 =0 => x=-3`TH2 :`(x+3)^2 -1 =0 => (x+3)^2 =1 => [(x+3=1),(x+3=-1):}``=> [(x=-2),(x=-4):}`Câu trả lời: `(x+3)^2014 = (x+3)^2012``(x+3)^2014 -(x+3)^2012 =0``(x+3)^2012 [(x+3)^2 -1]=0`TH1 :`(x+3)^2012 =0 => x+3 =0 => x=-3`TH2 :`(x+3)^2 -1 =0 => (x+3)^2 =1 => [(x+3=1),(x+3=-1):}``=> [(x=-2),(x=-4):}`

『Kuroba ム Tsuki Ryoo ︵²ᵏ⁷』 · Answer

`(x+3)^2014 = (x+3)^2012``=> (x+3)^2014 - (x+3)^2012 = 0``=> (x+3)^2 * (x+3)^2012 - (x+3)^2012 = 0``=> (x+3)^2012 * [ (x+3)^2 - 1] =0``=>`$\left[{}\begin{matrix}\left(x+3\right)^{2012}=0\\left(x+3\right)^2-1=0\end{matrix}\right.$`=>`$\left[{}\begin{matrix}x+3=0\\left(x+3\right)^2=1\end{matrix}\right.$`=>`$\left[{}\begin{matrix}x=-3\x+3=1\x+3=-1\end{matrix}\right.$`=>`$\left[{}\begin{matrix}x=-3\x=-2\x=-4\end{matrix}\right.$Vậy, `x = {-3; -2; -4}.`Câu trả lời: `(x+3)^2014 = (x+3)^2012``=> (x+3)^2014 - (x+3)^2012 = 0``=> (x+3)^2 * (x+3)^2012 - (x+3)^2012 = 0``=> (x+3)^2012 * [ (x+3)^2 - 1] =0``=>`$\left[{}\begin{matrix}\left(x+3\right)^{2012}=0\\left(x+3\right)^2-1=0\end{matrix}\right.$`=>`$\left[{}\begin{matrix}x+3=0\\left(x+3\right)^2=1\end{matrix}\right.$`=>`$\left[{}\begin{matrix}x=-3\x+3=1\x+3=-1\end{matrix}\right.$`=>`$\left[{}\begin{matrix}x=-3\x=-2\x=-4\end{matrix}\right.$Vậy, `x = {-3; -2; -4}.`

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