j) (2x²+1)(4x-3)=(x-12)(2x²+1)
⇔(2x²+1)(4x-3)-(x-12)(2x²+1)=0
⇔(2x²+1)(4x-3-x+12)=0
⇔\(\left[ \begin{array}{l}2x^2+1=0\\3x+9=0\end{array} \right.\)
⇔\(\left[ \begin{array}{l}x=\frac{-1}{2}\\x=-3\end{array} \right.\)
l)(x-1)(5x+3)=(3x-8)(x-1)
⇔(x-1)(5x+3)-(3x-8)(x-1)=0
⇔(x-1)(5x+3-3x+8)=0
⇔\(\left[ \begin{array}{l}x-1=0\\2x+11=0\end{array} \right.\)
⇔\(\left[ \begin{array}{l}x=1\\x=\frac{-11}{2}\end{array} \right.\)
n)(2-3x)(x+11)=(3x-2)(2-5x)
⇔-(3x-2)(x+11)-(3x-2)(2-5x)=0
⇔-(3x-2)(x+11+2-5x)=0
⇔\(\left[ \begin{array}{l}-3x+2=0\\-4x+13=0\end{array} \right.\)
⇔\(\left[ \begin{array}{l}x=\frac{2}{3}\\x=\frac{13}{4}\end{array} \right.\)
p)$(x-\frac{3}{4})^{2}$ + ($x-\frac{3}{4}$)($x-\frac{1}{2}$)=0
⇔($x-\frac{3}{4}$).$(x-\frac{3}{4}$ + $x-\frac{1}{2})$ = 0
⇔\(\left[ \begin{array}{l}x-\frac{3}{4}=0\\2x-\frac{5}{4}=0\end{array} \right.\)
⇔\(\left[ \begin{array}{l}x=\frac{3}{4}\\x=\frac{5}{8}\end{array} \right.\)
