x2 - 7x + 10 = 0
=>x2-2x=5x+10=0
=>x.(x-2)-5.(x-2)=0
=>(x-2)(x-5)=0
=>x=2 hoặc x=5
Đọc dc thì đọc
\(x^2-7x+10=0\)
\(\Rightarrow x^2-2x-5x+10=0\)
\(\Rightarrow x\left(x-2\right)-5\left(x-2\right)=0\)
\(\Rightarrow\left(x-2\right)\left(x-5\right)=0\)
\(\Leftrightarrow\left(x-2\right)=0\) \(\Leftrightarrow x=2\)
\(\left(x-5\right)=0\)\(\Leftrightarrow x=5\)
Vậy \(x\in\left\{2;5\right\}\)