a/ \(5\left(x+3\right)-2x\left(x+3\right)=0\)
\(\Leftrightarrow\left(x+3\right)\left(5-2x\right)=0\) \(\Leftrightarrow\left[\begin{array}{nghiempt}x=-3\\x=\frac{5}{2}\end{array}\right.\)
b/ \(4x\left(x-2004\right)-x+2004=0\)
\(\Leftrightarrow4x\left(x-2004\right)-\left(x-2004\right)=0\)
\(\Leftrightarrow\left(x-2007\right)\left(4x-1\right)=0\) \(\Leftrightarrow\left[\begin{array}{nghiempt}x=2007\\x=\frac{1}{4}\end{array}\right.\)
c/ \(\left(x+1\right)^2=x+1\Leftrightarrow\left(x+1\right)\left(x+1-1\right)=0\Leftrightarrow x\left(x+1\right)=0\)
\(\Leftrightarrow\left[\begin{array}{nghiempt}x=0\\x=-1\end{array}\right.\)
A) 5(x+3)-2x(3+x)=0
=> 5(x+3)-2x(x+3)=0
=> (5-2x)(x+3)=0
\(\Rightarrow\left[\begin{array}{nghiempt}5-2x=0\\x+3=0\end{array}\right.\)
\(\Rightarrow\left[\begin{array}{nghiempt}x=\frac{5}{2}\\x=-3\end{array}\right.\)
c)(x+1)2=x+1
=>(x+1)2-(x+1)=0
=>(x+1)(x+1-1)=0
=>(x+1)x=0
\(\Rightarrow\left[\begin{array}{nghiempt}x+1=0\\x=0\end{array}\right.\)
\(\Rightarrow\left[\begin{array}{nghiempt}x=-1\\x=0\end{array}\right.\)
