\(\left(x^3-x^2\right)-4x^2+8x-4=0\)
\(\Leftrightarrow x^3-x^2-4x^2+8x-4=0\)
\(\Leftrightarrow x^3-5x^2+8x-4=0\)
\(\Leftrightarrow x^3-x^2-4x^2+4x+4x-4=0\)
\(\Leftrightarrow x^2\left(x-1\right)-4x\left(x-1\right)+4\left(x-1\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left(x^2-4x+4\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left(x-2\right)^2=0\)
\(\Leftrightarrow\orbr{\begin{cases}x-1=0\\\left(x-2\right)^2=0\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}x=1\\x-2=0\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}x=1\\x=2\end{cases}}\)
Vậy ...
\(\left(x^3-x^2\right)-4x^2+8x-4=0\)
\(\Leftrightarrow x^3-x^2-4x^2+8x-4=0\)
\(\Leftrightarrow x^3-x^2-4x^2+4x+4x-4=0\)
\(\Leftrightarrow\left(x^3-x^2\right)-\left(4x^2+4x\right)+\left(4x-4\right)=0\)
\(\Leftrightarrow x^2\left(x-1\right)-4x\left(x-1\right)+4\left(x-1\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left(x^2-4x+4\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left(x-2\right)^2=0\)
\(\Leftrightarrow\orbr{\begin{cases}x-1=0\\x-2=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=1\\x=2\end{cases}}}\)
( x3 - x2 ) - 4x2 + 8x - 4 = 0
<=> x2( x - 1 ) - 4( x2 - 2x + 1 ) = 0
<=> x2( x - 1 ) - 4( x - 1 )2 = 0
<=> ( x - 1 )( x2 - 4x + 4 ) = 0
<=> ( x - 1 )( x - 2 )2 = 0
<=> x - 1 = 0 hoặc x - 2 = 0
<=> x = 1 hoặc x = 2
Vậy S = { 1 ; 2 }