a) x (x-2)+x-2=0
=> x.(x-2)+(x-2)=0
=> (x-2).(x+1)=0
=> Hoặc là x-2=0 => x=2
Hoặc là x+1=0 => x=-1
b) 5x (x-3)-x+3=0
=> 5x.(x-3)-(x-3)=0
=> (x-3).(5x-1)=0
=>Hoặc là x-3=0 => x=3
Hoặc là 5x-1=0 => x=1/5
a, x(x-2)+x-2=0
=> x(x-2)+(x-2)=0
=> (x-2).(x+1)=0
=> \(\hept{\begin{cases}x-2=0\\x+1=0\end{cases}}\Rightarrow\hept{\begin{cases}x=2\\x=-1\end{cases}}\)
b, 5x(x-3)-x+3=0
=> 5x(x-3)-(x-3)=0
=> (x-3).(5x-1)=0
=> \(\hept{\begin{cases}x-3=0\\5x-1=0\end{cases}}\Rightarrow\hept{\begin{cases}x=3\\x=\frac{1}{5}\end{cases}}\)
a) x ( x - 2 ) + x - 2 = 0
x ( x - 2 ) + ( x - 2 ) . 1= 0
( x - 2 ) ( x + 1 ) = 0
\(\Rightarrow\orbr{\begin{cases}x-2=0\\x+1=0\end{cases}}\Rightarrow\orbr{\begin{cases}x=2\\x=-1\end{cases}}\)
Vậy x = 2 ; x = -1
b) 5x ( x - 3 ) - x + 3 = 0
5x ( x - 3 ) - ( x - 3 ).1 = 0
( x - 3 ) ( 5x - 1 ) = 0
\(\Rightarrow\orbr{\begin{cases}x-3=0\\5x-1=0\end{cases}}\Rightarrow\orbr{\begin{cases}x=3\\5x=1\end{cases}}\Rightarrow\orbr{\begin{cases}x=3\\x=\frac{1}{5}\end{cases}}\)
Vậy x = 3 ; x = 1/5
a) \(x\left(x-2\right)+x-2=0\)
\(x\left(x-2\right)+\left(x-2\right)=0\)
\(\left(x+1\right)\left(x-2\right)=0\)
\(\Rightarrow\orbr{\begin{cases}x+1=0\\x-2=0\end{cases}}\Rightarrow\orbr{\begin{cases}x=-1\\x=2\end{cases}}\)
vay \(\orbr{\begin{cases}x=-1\\x=2\end{cases}}\)
b) \(5x\left(x-3\right)-x+3=0\)
\(5x\left(x-3\right)-\left(x-3\right)=0\)
\(\left(5x-1\right)\left(x-3\right)=0\)
\(\Rightarrow\orbr{\begin{cases}5x-1=0\\x-3=0\end{cases}}\Rightarrow\orbr{\begin{cases}x=\frac{1}{5}\\x=3\end{cases}}\)
vay \(\orbr{\begin{cases}x=\frac{1}{5}\\x=3\end{cases}}\)
