a) \(x^2-x+x-1-x^2+2x=3\)
(=) \(-1+2x=3\)
(=) \(2x=3+1\)
(=) \(2x=4\)
(=) \(x=4:2\)
(=) \(x=2\)
b) \(x^2-x-3x+3=0\)
(=) \(x\left(x-1\right)-3\left(x-1\right)=0\)
(=) \(\left(x-1\right)\left(x-3\right)=0\)
(=) \(\orbr{\begin{cases}x-1=0\\x-3=0\end{cases}}\)(=) \(\orbr{\begin{cases}x=0+1\\x=0+3\end{cases}}\)(=) \(\orbr{\begin{cases}x=1\\x=3\end{cases}}\)
a.)(x+1)(x-1)-x(x+2)=3
x^2-1-x^2-2x=3
-1-2x=3
x=-2
b.)x^2-4x+3=0
x^2-3x-x+3=0
x(x-3)-(x-3)=0
(x-3)(x-1)=0
suy ra x-3=0 hoặc x-1=0
x=3 hoặc x=1
a, (x+1)(x-1)-x(x+2)=3
x2-1-x2-2x=3
-1-2x=3
-2x=-4
x=2
b, x2-4x+3=0
x2-3x-x+3=0
x2-x-3x+3=0
x(x-1)-3(x-1)=0
(x-1)(x-3)=0
=>x-1=0=>x=1
=>x-3=0=>x=3
