Câu hỏi của Otasaka Yu

Question

$\left(x-2\right)^4+\left(x-3\right)^4=1$

Edogawa Conan · Accepted Answer

Đặt y = x - 2

Ta có:

$y^4+\left(y-1\right)^4=1$

$\Leftrightarrow y^4+y^4-4y^3+6y^2-4y+1=1$

$\Leftrightarrow2y^4-4y^3+6y^2-4y=0$

$\Leftrightarrow y\left(y-1\right)\left(2y^2-2y+4\right)$Câu trả lời: Đặt y = x - 2

Ta có:

$y^4+\left(y-1\right)^4=1$

$\Leftrightarrow y^4+y^4-4y^3+6y^2-4y+1=1$

$\Leftrightarrow2y^4-4y^3+6y^2-4y=0$

$\Leftrightarrow y\left(y-1\right)\left(2y^2-2y+4\right)$

Edogawa Conan · Answer

nãy bấm nhầm

Đặt y = x - 2

ta có:

$y^4+\left(y-1\right)^4=1$

$\Leftrightarrow y^4+y^4-4y^3+6y^2-4y+1=1$

$\Leftrightarrow2y^4-4y^3+6y^2-4y=0$

$\Leftrightarrow y\left(y-1\right)\left(2y^2-2y+4\right)=0$

Mà: 2y2 -2y + 4 >0

$\Rightarrow\left[{}\begin{matrix}y=0\y-1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}y=0\y=1\end{matrix}\right.$

$\Rightarrow\left[{}\begin{matrix}x-2=0\x-2=1\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=2\x=3\end{matrix}\right.$Câu trả lời: nãy bấm nhầm

Đặt y = x - 2

ta có:

$y^4+\left(y-1\right)^4=1$

$\Leftrightarrow y^4+y^4-4y^3+6y^2-4y+1=1$

$\Leftrightarrow2y^4-4y^3+6y^2-4y=0$

$\Leftrightarrow y\left(y-1\right)\left(2y^2-2y+4\right)=0$

Mà: 2y2 -2y + 4 >0

$\Rightarrow\left[{}\begin{matrix}y=0\y-1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}y=0\y=1\end{matrix}\right.$

$\Rightarrow\left[{}\begin{matrix}x-2=0\x-2=1\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=2\x=3\end{matrix}\right.$

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