\(\text{a) 3.(x-2)+x.(x-2)=0}\)
\(\Leftrightarrow\)\(\text{(x-2)(3+x)=0}\)
\(\Leftrightarrow\left[\begin{array}{nghiempt}x-2=0\\3+x=0\end{array}\right.\) \(\Leftrightarrow\left[\begin{array}{nghiempt}x=2\\x=-3\end{array}\right.\)
\(\text{Vậy x=2 hoặc x=-3}\)
\(b,4x.\left(x-2\right)-x+2\)=0
\(\Leftrightarrow4x.\left(x-2\right)-\left(x-2\right)\)=0
\(\Leftrightarrow\left(x-2\right)\left(4x-1\right)\)=0
\(\Leftrightarrow\left[\begin{array}{nghiempt}x-2=0\\4x-1=0\end{array}\right.\)
\(\Leftrightarrow\left[\begin{array}{nghiempt}x=2\\x=\frac{1}{4}\end{array}\right.\)
Vậy x=2 hoặc \(x=\frac{1}{4}\)
a) 3.(x-2) + x. ( x-2) = 0
(x - 2)(3 + x) = 0
TH1:
x - 2 = 0
x = 2
TH2:
3 + x = 0
x = -3
Vậy x = 2 hoặc x = -3
b) 4x.(x-2) -x +2 = 0
4x(x - 2) - x + 2 = 0
(x - 2)(4x - 1) = 0
TH1:
x - 2 = 0
x = 2
4x - 1 = 0
4x = 1
x = 1/4
Vậy x = 2 hoặc x = 1/4
a) 3 . ( x - 2 ) + x . ( x - 2 ) = 0
=> 3x - 6 + 2x - 2x = 0
=> 3x + 2x - 2x = 0 + 6
=> 3x = 6
=> x = 6 : 3 = 2
b) 4x . ( x - 2 ) - x + 2 = 0
=> 5x - 6x - x + 2 = 0
=> 5x - 6x - x = 0 - 2 = - 2
=> - 2x = - 2
=> x = - 2 : ( - 2 )
=> x = 1
