Câu hỏi của Phạm Kiên

Question

(x-3)mu7bằng(x-3)mũ2.Giải giúp mình với

The Lonely Cancer · Accepted Answer

$\left(x-3\right)^7=\left(x-3\right)^2\Leftrightarrow\left(x-3\right)^7-\left(x-3\right)^2=0$

$\Leftrightarrow\left(x-3\right)^2\left[\left(x-3\right)^7-1\right]=0$

$\Rightarrow\orbr{\begin{cases}\left(x-3\right)^2=0\\left(x-3\right)^7-1=0\end{cases}}$

$\Rightarrow\orbr{\begin{cases}x-3=0\\left(x-3\right)^7=1\end{cases}}$

$\Rightarrow\orbr{\begin{cases}x=0+3=3\x-3=1\Rightarrow x=4\end{cases}}$

Vậy $x\in\left\{3;4\right\}$

Câu trả lời:

$\left(x-3\right)^7=\left(x-3\right)^2\Leftrightarrow\left(x-3\right)^7-\left(x-3\right)^2=0$

$\Leftrightarrow\left(x-3\right)^2\left[\left(x-3\right)^7-1\right]=0$

$\Rightarrow\orbr{\begin{cases}\left(x-3\right)^2=0\\left(x-3\right)^7-1=0\end{cases}}$

$\Rightarrow\orbr{\begin{cases}x-3=0\\left(x-3\right)^7=1\end{cases}}$

$\Rightarrow\orbr{\begin{cases}x=0+3=3\x-3=1\Rightarrow x=4\end{cases}}$

Vậy $x\in\left\{3;4\right\}$