a. 3x(x-2)-x+2=0
3x(x-2)-(x-2)=0
(3x-1)(x-2)=0
=>\(\hept{\begin{cases}3x-1=0\\x-2=0\end{cases}}\)
=> \(\hept{\begin{cases}3x=1\\x=2\end{cases}}\)
=>\(\hept{\begin{cases}x=\frac{1}{3}\\x=2\end{cases}}\)
vậy x thuộc (1/3;2)
b. 4x(x-3)-2x+6=0
4x(x-3) -2(x-3)=0
(4x-2)(x-3)
=>*4x-2=0
4x=2
x=1/2
*x-3=0
x=3
vậy x thuộc (1/2;3)
a. 3x(x-2)-x+2=0
=>(x-2)(3x-1)=0
=>x-2=0 hoặc 3x-1=0
*x-2=0=>x=2
*3x-1=0=>x=\(\frac{1}{3}\)
Vậy x=2;x=\(\frac{1}{3}\)
b. 4x(x-3)-2x+6=0
=>(x-3)(4x-2)=0
=>(x-3)=0 hoặc 4x-2=0
*x-3=0=>x=3
*4x-2=0=>\(\frac{1}{2}\)
Vậy x=3; x=\(\frac{1}{2}\)
câu c, d,e thì chút nx mk gửi cho nha
a, 3x ( x - 2 ) - x + 2 = 0
=> \(3x^2-6x-x+2=0\)
=> \(3x^2-7x+2=0\)
=> \(\left(x-2\right)\left(3x-1\right)=0\)
=> \(\orbr{\begin{cases}x-2=0\\3x-1=0\end{cases}\Rightarrow\orbr{\begin{cases}x=2\\x=\frac{1}{3}\end{cases}}}\)
b, 4x ( x - 3 ) - 2x + 6 = 0
=> \(4x^2-12x-2x+6=0\)
=> \(4x^2-14x+6=0\)
=> \(\left(x-3\right)\left(2x-1\right)=0\)
=> \(\orbr{\begin{cases}x-3=0\\2x-1=0\end{cases}\Rightarrow\orbr{\begin{cases}x=3\\x=\frac{1}{2}\end{cases}}}\)
c, 2x ( x - 4 ) + x - 4 = 0
=> \(2x^2-8x+x-4=0\)
=> \(2x^2-7x-4=0\)
=> \(\left(x-4\right)\left(2x+1\right)=0\)
=> \(\orbr{\begin{cases}x-4=0\\2x+1=0\end{cases}\Rightarrow\orbr{\begin{cases}x=4\\x=-\frac{1}{2}\end{cases}}}\)
d, \(2x^3+4x=0\)
=> \(2x\left(x^2+2\right)=0\)
=> \(\orbr{\begin{cases}2x=0\\x^2+2=0\end{cases}\Rightarrow\orbr{\begin{cases}x=0\\x^2=-2\end{cases}}}\)( vô lý ) => x = 0
Vậy x = 0
e, \(3x^3-6x=0\)
=> \(3x\left(x^2-2\right)=0\)
=> \(\orbr{\begin{cases}3x=0\\x^2-2=0\end{cases}\Rightarrow\orbr{\begin{cases}x=0\\x^2=2\end{cases}}}\Rightarrow[\begin{cases}x=0\\x=\sqrt{2}\\x=-\sqrt{2}\end{cases}\)
