\(a,7x+3x=2010\\ x\left(7+3\right)=2010\\ 10x=2010\\ x=201\)
Vậy x = 201
\(b,\left(x-6\right)\left(x-12\right)=0\\ \Rightarrow\left[{}\begin{matrix}x-6=0\\x-12=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=6\\x=12\end{matrix}\right.\)
Vậy \(x\in\left\{6;12\right\}\)
a. 7.x + 3.x=2010
=> 10.x = 2010
=> x = 201
b. (x-6).(x-12)=0
<=> x-6 = 0
x-12 = 0
<=> x = 6
x = 12
a) 7.x+3.x=2010
⇔ (7 + 3) . x = 2010
⇔ 10 . x = 2010
⇔ x = 201
b) (x-6).(x-12)=0
⇔ x - 6 = 0 hoặc x - 12 = 0
⇔ x ∈ { 6;12 }
a) 7.x + 3.x = 2010
(7 + 3) .x = 2010
10 . x = 2010
x = 2010 : 10
x = 201
b) ( x - 6 ) . ( x - 12 ) = 0
TH1 : x - 6 = 0 TH2 : x - 12 = 0
x = 0 + 6 x = 0 + 12
x = 6 x = 12
Vậy x ∈ { 6 ; 12 }
