\(\left(x-2\right)^2-5x+10=0\)
\(\Leftrightarrow\left(x-2\right)^2+5\left(x-2\right)=0\)
\(\Leftrightarrow\left(x-2\right)\left(x+3\right)=0\)
\(\left\{{}\begin{matrix}x-2=0\\x+3=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=2\\x=-3\end{matrix}\right.\)
Vậy \(x=2\) hoặc \(x=-3\)
(x - 2)2 - 5x + 10 = 0
\(\Rightarrow\) x2 - 4x + 4 - 5x = -10
\(\Rightarrow\) x2 - 9x = -14
\(\Rightarrow\) x2 - 9x = 72 - 9 . 7
\(\Rightarrow\) x = 7
(x-2)2-5x+10=0
=>x2-4x+4-5x+10=0
=>x2-9x+14=0
=>x2-2x-7x+14=0
=>(x2-2x)-(7x-14)=0
=>x(x-2)-7(x-2)=0
=>(x-2)(x-7)=0
=>\(\left[{}\begin{matrix}x-2=0\\x-7=0\end{matrix}\right.\) =>\(\left[{}\begin{matrix}x=2\\x=7\end{matrix}\right.\)
