Question

(x-1)^2 =(x-1)^4
giúp mình với 

Answer 1

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

\(\Rightarrow\left(x-1\right)^4-\left(x-1\right)^2=0\)

\(\Rightarrow\left(x-1\right)^2\left[\left(x-1\right)^2-1\right]=0\)

\(\Rightarrow\left[{}\begin{matrix}x-1=0\\\left(x-1\right)^2=1\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}x=1\\x-1=1\\x-1=-1\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}x=1\\x=2\\x=0\end{matrix}\right.\)

Answer 2

\(\Rightarrow\left[{}\begin{matrix}x=0\\x=1\end{matrix}\right.\)

 

Answer 3

<=> (x - 1)^2 = 1

<=> x - 1 = 1

<=> x = 2

Answer 4

`(x-1)^2 = (x-1)^4`

`=> (x-1)^4 - (x-1)^2 =0`

`=> (x-1)^2 [ (x-1)^2 - 1] =0`

`=> `\(\left[ \begin{array}{l}(x-1)^2=0\\(x-1)^2-1=0\end{array} \right.\) 

`=>` \(\left[ \begin{array}{l}x-1=0\\(x-1)^2=1\end{array} \right.\) 

`=>` \(\left[ \begin{array}{l}x=1\\x-1=1\\x-1=-1\end{array} \right.\) 

`=>` \(\left[ \begin{array}{l}x=1\\x=2\\x=0\end{array} \right.\) 

Vậy `x in { 1; 2;0}`