a) $(2x+5)(x-4)=(x-4)(5-x)$
$⇔(2x+5)(x-4)-(x-4)(5-x)=0$
$⇔(x-4)(2x+5-5+x)=0$
$⇔(x-4)3x=0$
⇔\(\left[ \begin{array}{l}x-4=0\\3x=0\end{array} \right.\)
⇔\(\left[ \begin{array}{l}x=4\\x=0\end{array} \right.\)
b) $16x²-25=(4x-5)(2x+1)$
$⇔(4x-5)(4x+5)-(4x-5)(2x+1)=0$
$⇔(4x-5)(4x+5-2x-1)=0$
$⇔(4x-5)(2x-4)=0$
⇔\(\left[ \begin{array}{l}4x-5=0\\2x-4=0\end{array} \right.\)
⇔ \(\left[ \begin{array}{l}x=\frac{5}{4}\\x=2\end{array} \right.\)
c) $(3x+1)²-4(x-3)²=0$
$⇔(9x²+6x+1)-4(x²-6x+9)=0$
$⇔9x²+6x+1-4x²+24x-36=0$
$⇔5x²+30x-35=0$
$⇔5x²-5x+35x-35=0$
$⇔5x(x-1)+35(x-1)=0$
$⇔(x-1)(5x-35)=0$
⇔\(\left[ \begin{array}{l}x-1=0\\5x-35=0\end{array} \right.\)
⇔\(\left[ \begin{array}{l}x=1\\x=7\end{array} \right.\)
