a) $2x(x-3)-(3-x)$=$0$
=>$2x(x-3)-(x-3)$=$0$
=>$(2x-1)(x-3)$=$0$
=>\(\left[ \begin{array}{l}2x-1=0\\x-3=0\end{array} \right.\)
=>\(\left[ \begin{array}{l}2x=1\\x=3\end{array} \right.\)
=>\(\left[ \begin{array}{l}x=\frac{1}{2}\\x=3\end{array} \right.\)
Vậy phương trình có 2 nghiệm: $x1$= $\frac{1}{2}$ ; $x$=3
b) $3x(x+5)-6(x+5)$=$0$
=>$(3x-6)(x+5)$=$0$
=>\(\left[ \begin{array}{l}3x-6=0\\x+5=0\end{array} \right.\)
=>\(\left[ \begin{array}{l}3x=6\\x=-5\end{array} \right.\)
=>\(\left[ \begin{array}{l}x=2\\x=-5\end{array} \right.\)
c) $x^{4}$ -$x^{2}$ =$0$
=>$x^{2} (x^{2} -1)$=$0$
=>\(\left[ \begin{array}{l}x^{2}=0\\x^{2}-1=0\end{array} \right.\)
=>\(\left[ \begin{array}{l}x=0\\x=1\end{array} \right.\)
Vậy phương trình trên có 2 nghiệm là: $x1$=0 ; $x2$=1
d)
$x^{n}(x+1)-$ $x^{n}$ -$x^{n-1}$
=>$x^{n}(x+1)$ -($x^{n}$ -$x^{n-1 }$)
=>$x^{n}(x+1)$ -($x^{n}$ +$x^{n }.x$)
=>$x^{n}(x+1)$ -$x^{n}(1+x$)
=>$x^{n}(x+1)$ -$x^{n}(x+1$)
=>($x^{n}$ -$x^{n}$)$(x+1)$ =$0$
=>\(\left[ \begin{array}{l}x^{n}-x^{n}=0\\x+1=0\end{array} \right.\)
=> \(\left[ \begin{array}{l}x=0\\x=-1\end{array} \right.\)
