(x-2)(x-6)=0
x-2=0 hoặc x-6=0
x=2 hoắc x=6
\(\Rightarrow x=2\)
Chỉ pt tới dok thuj!^^
\(x^2-8x+12=0\)
=>\(x^2-2x-6x+12=0\)
=>\(x\left(x-2\right)-6\left(x-2\right)=0\)
=>\(\left(x-6\right)\left(x-2\right)=0\)
=>\(\orbr{\begin{cases}x-6=0\\x-2=0\end{cases}\Rightarrow\orbr{\begin{cases}x=6\\x=2\end{cases}}}\)
Vậy ...
Ta có\(x^2-8x+12=\left(x^2-8x+16\right)-4\)
\(=\left(x-4\right)^2=4\)\(\Rightarrow\orbr{\begin{cases}x=6\\x=2\end{cases}}\)
k nha
=> x2-6x-2x+12=0
=> (x2-6x)- (2x-12) =0
=> x(x-6) - 2(x-6) = 0
=> (x-6)(x-2) =0
=> Xảy ra 2 TH:
TH1: x-6=0 TH2: x-2 = 0
<=> x = 6 <=> x= 2
Vậy X=6 ; X=2