Câu hỏi của Vu Hai Anh

Question

Tìm x :

$_{\left(2-x\right)^8+\left(2-x\right)^9=0}$

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Phạm Thị Thùy Linh · Accepted Answer

$\left(x-2\right)^8+\left(2-x\right)^9=0.$

$\Rightarrow\left(2-x\right)^8\left(1+2-x\right)=0$

$\Rightarrow\left(2-x\right)^8\left(3-x\right)=0$

$\Rightarrow x\in\left\{2;3\right\}$

Câu trả lời:

$\left(x-2\right)^8+\left(2-x\right)^9=0.$

$\Rightarrow\left(2-x\right)^8\left(1+2-x\right)=0$

$\Rightarrow\left(2-x\right)^8\left(3-x\right)=0$

$\Rightarrow x\in\left\{2;3\right\}$

Xyz OLM · Answer

(2 - x)⁸ + (2 - x)⁹ = 0

=> (2 - x)⁸. (1 + 2 - x) = 0

=> (2 - x)⁸ . (3 - x) = 0

=> $\orbr{\begin{cases}\left(2-x\right)^8=0\3-x=0\end{cases}}$

=> $\orbr{\begin{cases}\left(2-x\right)^8=0^8\x=3-0\end{cases}}$

=> $\orbr{\begin{cases}2-x=0\x=3\end{cases}}$

=> $\orbr{\begin{cases}x=2-0\x=3\end{cases}}$

=> $\orbr{\begin{cases}x=2\x=3\end{cases}}$

Câu trả lời:

(2 - x)⁸ + (2 - x)⁹ = 0

=> (2 - x)⁸. (1 + 2 - x) = 0

=> (2 - x)⁸ . (3 - x) = 0

=> $\orbr{\begin{cases}\left(2-x\right)^8=0\3-x=0\end{cases}}$

=> $\orbr{\begin{cases}\left(2-x\right)^8=0^8\x=3-0\end{cases}}$

=> $\orbr{\begin{cases}2-x=0\x=3\end{cases}}$

=> $\orbr{\begin{cases}x=2-0\x=3\end{cases}}$

=> $\orbr{\begin{cases}x=2\x=3\end{cases}}$

Hoàng Long · Answer

$\left(2-x\right)^8+\left(2-x\right)^9=0$

$=>\left(2-x\right)^8\left(3-x\right)=0$

$=>x\in\left\{2;3\right\}$

Câu trả lời:

$\left(2-x\right)^8+\left(2-x\right)^9=0$

$=>\left(2-x\right)^8\left(3-x\right)=0$

$=>x\in\left\{2;3\right\}$