\(\Rightarrow x^2-x-5x+5=0\)
\(\Rightarrow\left(x^2-x\right)-\left(5x-5\right)=0\)
\(\Rightarrow x.\left(x-1\right)-5.\left(x-1\right)=0\)
\(\Rightarrow\left(x-1\right).\left(x-5\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x-1=0\\x-5=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=1\\x=5\end{matrix}\right.\)
\(\Leftrightarrow x^2-x-5x+5=0\\ \Leftrightarrow x\left(x-1\right)-5\left(x-1\right)=0\\ \Leftrightarrow\left(x-5\right)\left(x-1\right)=0\Leftrightarrow\left[{}\begin{matrix}x=1\\x=5\end{matrix}\right.\)
Pt có dạng a + b + c = 1 + (-6) + 5 = 0
Nên ta có
x1 = 1
x2 = c/a = 5
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