Đáp án:
$a,x^2-4x-21=0^{}$
$x^{2}+3x-7x-21=0$
$x(x^{}+3)-7(x+3)=0$
$(x+3).(x-7)^{}=0$
\(⇔\left[ \begin{array}{l}x+3=0\\x-7=0\end{array} \right.\)
\(⇔\left[ \begin{array}{l}x=-3\\x=7\end{array} \right.\)
$Vậy $ $x=-3$ $hoặc$ $x=7$
$b,\frac{ 3x+1}{x^2-9}=$ $\frac{2}{x-3}-$ $\frac{x-3}{x+3}$
$\frac{3x+1}{(x-3)(x+3)}=$ $\frac{2.(x+3)}{(x-3)(x+3)}-$ $\frac{(x-3)^2}{(x-3).(x+3)}$
$\frac{3x+1}{(x-3)(x+3)}=$ $\frac{2x+6}{(x-3)(x+3)}-$ $\frac{x^2-6x+9}{(x-3)(x+3)}$
$\frac{3x+1}{(x-3)(x+3)}=$ $\frac{2x+6-x^2+6x-9}{(x-3)(x+3)}$
$⇔3x+1=2x+6-x^2+6x-9$
$⇔x^2+3x-8x+1+3=0$
$⇔x^2-5x+4$
$⇔x^{}(x-4)-(x-4)=0$
$⇔(x^{}-4).(x-1)$
\(⇔\left[ \begin{array}{l}x=4\\x=1\end{array} \right.\)
$Vậy $ $x=4$ $hoặc$ $x=1$
c, $/2x+1/=x-7$
$ĐKXD:x^{}$ $\geq7$
\(⇔\left[ \begin{array}{l}2x+1=x-7\\2x+1=7-x\end{array} \right.\)
\(\left[ \begin{array}{l}x=-8(l)\\x=2(l)\end{array} \right.\)
$Vậy$ $phương^{}$ $trình$ $vô$ $nghiệm^{}$
