Mình phân tích ở trên rồi
Ta có: \(\left(x-1\right)\left(x-2\right)\left(x-3\right)=0\)
=> x - 1 = 0 => x = 0+1 => x = 1
Hoặc x - 2 = 0 => x = 0+2 => x = 2
Hoặc x - 3 = 0 => x = 0+3 => x = 3
Vậy x = 1 hoặc x = 2 hoặc x = 3
\(\left(x-1\right)\left(x^2-5x+6\right)=\left(x-1\right)\left(x^2-2x-3x+6\right)\)
\(=\left(x-1\right)\left[x\left(x-2\right)-3\left(x-2\right)\right]\)
\(=\left(x-1\right)\left(x-2\right)\left(x-3\right)\)
\(\left(x-1\right)\left(x^2-5x+6\right)=0\)
\(\Leftrightarrow\left(x-1\right).\left(x-2\right).\left(x-3\right)=0\)
\(\Leftrightarrow\hept{\begin{cases}x-1=0\\x-2=0\\x-3=0\end{cases}}\)
\(\Leftrightarrow\hept{\begin{cases}x=1\\x=2\\x=3\end{cases}}\)
Vậy .......................
