\(\left(x-1\right)\left(x-2\right)\left(x-3\right)\left(x-4\right)-\left(x-2\right)\left(x-4\right)\left(x-5\right)=0\)
\(\Rightarrow\left(x-2\right)\left(x-4\right)\left[\left(x-1\right)\left(x-3\right)-\left(x-5\right)\right]=0\)
\(\Rightarrow\left(x-2\right)\left(x-4\right)\left(x^2-4x+3-x+5\right)=0\)
\(\Rightarrow\left(x-2\right)\left(x-4\right)\left(x^2-5x+8\right)=0\)
\(\Rightarrow\orbr{\begin{cases}x-2=0\\x-4=0\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}x=2\\x=4\end{cases}}\)
Vậy tổng là : 6
\(\left(x+1\right)\left(x-2\right)\left(x-3\right)\left(x-4\right)-\left(x-2\right)\left(x-4\right)\left(x-5\right)=0\)
\(\Leftrightarrow\left(x-2\right)\left(x+4\right)\left[\left(x-1\right)\left(x-3\right)-\left(x-5\right)\right]=0\)
\(\Leftrightarrow\left(x-2\right)\left(x-4\right)\left(x^2-4x+3-x+5\right)=0\)
\(\Leftrightarrow\left(x-2\right)\left(x-4\right)\left(x^2-5x+8\right)=0\)
\(\Leftrightarrow\hept{\begin{cases}x-2=0\\x-4=0\\x^2-5x+8=0\end{cases}}\)
\(\Leftrightarrow\hept{\begin{cases}x=2\\x=4\\x^2-5x+8=0->ktm\end{cases}}\)
\(x^2-5x+8=0\Leftrightarrow\left(x-\frac{5}{2}\right)^2+\frac{7}{4}\ge\frac{7}{4}>0=>ktm\)
cn lại tự lm nha bn