a) (x-4)(3x+1)=(x-4)(x+1) <=> (x-4)(3x+1-x-1) =0 <=>2x.(x-4)=0 <=> x=0 hoặc x=4 b) $x^{2}$ -9+4x(x-3)=0 <=>$x^{2}$ -9+4$x^{2}$ -12x=0 <=>5$x^{2}$ -12x-9=0 <=>5$x^{2}$ -15x+3x-9=0 <=> (x-3)(5x+3)=0 <=> \(\left[ \begin{array}{l}x=3\\x=\frac{-3}{5}\end{array} \right.\) c)$(2x-1)^{2}$ =$(x-3)^{2}$ <=>$(2x-1)^{2}$ -$(x-3)^{2}$=0 <=> (2x+1-x+3)(2x+1+x-3)=0 <=>(x+4)(3x-2)=0 <=>$\left \{ {{x=\frac{2}{3}} \atop {x=-4}} \right.$ d)4$x^{2}$ -25=(5-2x)(4x+3) <=> 4$x^{2}$ -25=20x+15-8$x^{2}$ -6x <=>12$x^{2}$ -14x-40=0 <=>6$x^{2}$ -7x-20=0 <=> (2x-5)(3x+4)=0 \(\left[ \begin{array}{l}x=\frac{5}{2}\\x=\frac{-4}{3}\end{array} \right.\) e)($x^{2}$ +x-1)($x^{2}$ +x+3)=5 Đặt $x^{2}$ +x=a => (a-1)(a+3)=5 <=>$a^{2}$ +2a-8=0 <=> \(\left[ \begin{array}{l}a=2\\a=-4\end{array} \right.\) <=>\(\left[ \begin{array}{l}x^2+x-2=0\\x^2+x+4=0(vô nghiệm)\end{array} \right.\) <=>\(\left[ \begin{array}{l}x=1\\x=-2\end{array} \right.\) f)$\frac{x+3}{x-3}$ +$\frac{x-3}{x+3}$= $\frac{x^2+2x}{x^2-9}$ +1 (ĐK x $\neq$ $\pm$3) <=>$\frac{x^2+6x+9+x^2-6x+9}{(x-3)(x+3)}$ =$\frac{x^2+2x+x^2-9}{(x-3)(x+3)}$ =>2$x^{2}$ +18=2$x^{2}$ +2x-9 <=>2x=27 <=>x=$\frac{27}{2}$ (t/m) g) 5+$\frac{96}{x^2-16}$ =$\frac{2x-1}{x+4}$ +$\frac{3x-1}{x-4}$ (ĐK x$\neq$ $\pm$4) <=>$\frac{5x^2-80+96}{(x-4)(x+4)}$ =$\frac{2x^2-9x+4+3x^2+11x-4}{(x+4)(x-4)}$ =>5$x^{2}$ +16=5$x^{2}$ +2x <=>2x=16 <=>x=8 (t/m) h)$\frac{x}{2x-6}$ +$\frac{x}{2x+2}$ =$\frac{2x+4}{x^2-2x-3}$ (ĐK x$\neq$3,x$\neq$-1) <=>$\frac{x^2+x+x^2-3x}{2(x-3)(x+1)}$ =$\frac{4x+8}{2(x-3)(x+1)}$ => 2$x^{2}$ -2x=4x+8 <=>2$x^{2}$ -6x-8=0 <=>$x^{2}$ -3x-4=0 \(\left[ \begin{array}{l}x=4\\x=-1(l)\end{array} \right.\) \(\left[ \begin{array}{l}x=2\\x=-1\end{array} \right.\)