a) \(3x\left(x-2\right)-x+2=0\)
\(3x\left(x-2\right)-\left(x-2\right)=0\)
\(\left(x-2\right)\left(3x-1\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x-2=0\\3x-1=0\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=2\\x=\dfrac{1}{3}\end{matrix}\right.\)
b) \(x^2+5x+4=0\)
\(x^2+x+4x+4=0\)
\(x\left(x+1\right)+4\left(x+1\right)=0\)
\(\left(x+1\right)\left(x+4\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x+1=0\\x+4=0\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=-1\\x=-4\end{matrix}\right.\)
\(a,3x\left(x-2\right)-x+2=0\)
\(\Rightarrow3x\left(x-2\right)-\left(x-2\right)=0\)
\(\Rightarrow\left(x-2\right)\left(3x-1\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x-2=0\\3x-1=0\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=2\\x=\dfrac{1}{3}\end{matrix}\right.\)
\(b,x^2+5x+4=0\)
\(\Rightarrow x^2+4x+x+4=0\)
\(\Rightarrow\left(x^2+4x\right)+\left(x+4\right)=0\)
\(\Rightarrow x\left(x+4\right)+\left(x+4\right)=0\)
\(\Rightarrow\left(x+4\right)\left(x+1\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x+4=0\\x+1=0\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=-4\\x=-1\end{matrix}\right.\)
a) \(3x\left(x-2\right)-x+2=0\)
\(\Rightarrow3x\left(x-2\right)-\left(x-2\right)=0\)
\(\Rightarrow\left(x-2\right)\left(3x-1\right)=0\)
\(\Rightarrow\left\{{}\begin{matrix}x-2=0\\3x-1=0\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}x=2\\x=\dfrac{1}{3}\end{matrix}\right.\)
Vậy ...
b) \(x^2+5x+4=0\)
\(\Rightarrow x^2+4x+x+4=0\)
\(\Rightarrow x\left(x+4\right)+\left(x+4\right)=0\)
\(\Rightarrow\left(x+4\right)\left(x+1\right)=0\)
\(\Rightarrow\left\{{}\begin{matrix}x+4=0\\x+1=0\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}x=-4\\x==-1\end{matrix}\right.\)
Vậy ...
a) 3x ( x - 2 ) - x + 2 = 0
=> 3x ( x - 2 ) - ( x - 2) = 0
=> ( x - 2 ) ( 3x - 1 ) = 0
=> x - 2 = 0
3x - 1 = 0
=> x = 2
x = \(\dfrac{1}{3}\)
b) \(x^2+5x+4\)= 0
=> \(x^2+x+4x+4=0\)
=> x ( x + 1 ) + 4 ( x + 1 ) = 0
=> ( x + 1 ) ( x + 4 ) = 0
=> x + 1 = 0
x + 4 = 0
=> x = -1
x = -4
