Question

Tìm \(x\in Q\), biết:

   \(x\left(6-x\right)^{2003}=\left(6-x\right)^{2003}\)

Answer 1

TH1: \(6-x=0\)

\(\Rightarrow x=6-0=6\)

TH2: \(6-x\ne0\)

\(\Rightarrow x=\frac{\left(6-x\right)^{2003}}{\left(6-x\right)^{2003}}=1\)

Vậy \(\orbr{\begin{cases}x=6\\x=1\end{cases}}\)

Answer 2

x = 6 và x = 1

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Answer 3

chuan

Answer 4

+) \(x\left(6-x\right)^{2003}=\left(6-x\right)^{2003}\Rightarrow x=\left(6-x\right)^{2003}:\left(6-x\right)^{2003}\Rightarrow x=1\)

+)\(x\left(6-x\right)^{2003}=\left(6-x\right)^{2003}\Rightarrow x\left(6-x\right)^{2003}-\left(6-x\right)^{2003}=0\Rightarrow x=6\)

vậy......

Answer 5

\(x\left(6-x\right)^{2003}=\left(6-x\right)^{2003}\)

=> \(x\left(6-x\right)^{2003}-\left(6-x\right)^{2003}=0\)

=>\(\left(6-x\right)^{2003}.\left(x-1\right)=0\)

=> \(\text{Hoặc}\left(6-x\right)^{2003}=0\text{Hoặc}\left(x-1\right)=0\)

TH1: \(\left(6-x\right)^{2003}=0\)

=> \(6-x=0\)

=>\(x=6\)

TH2:\(x-1=0\)

=>\(x=1\)

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