Question

Tìm x biết:

2006.|x-1| + (x-1)² = 2005. |1-x|

Answer 1

Ta có :

\(2006\left|x-1\right|+\left(x-1\right)^2=2005\left|1-x\right|\)

\(\Rightarrow2006\left|x-1\right|+\left(x-1\right)^2=2005\left|x-1\right|\)

\(\Rightarrow2006\left|x-1\right|+\left(x-1\right)^2-2005\left|x-1\right|=0\)

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

Vì \(\begin{cases}\left|x-1\right|\ge0\\\left(x-1\right)^2\ge0\end{cases}\)\(\forall x\)

\(\Rightarrow\begin{cases}x-1=0\\x-1=0\end{cases}\)

=> x = 1

Vậy x = 1