a, 3(x-1)(2x-1)=5(x+8)(x-1)
⇔3(x-1)(2x-1)-5(x+8)(x-1)=0
⇔(x-1)[3(2x-1)-5(x+8)]=0
⇔(x-1)(6x-3-5x-40)=0
⇔(x-1)(x-43)=0
⇔\(\left[ \begin{array}{l}x-1=0\\x-43=0\end{array} \right.\) <=>\(\left[ \begin{array}{l}x=1\\x=43\end{array} \right.\)
Vậy S={1;43}
b, 9x²-1=(3x-1)(4x+1)
⇔(3x-1)(3x+1)-(3x-1)(4x+1)=0
⇔(3x-1)(3x+1-4x-1)=0
⇔(3x-1).(-x)=0
⇔\(\left[ \begin{array}{l}3x-1=0\\-x=0\end{array} \right.\) <=>\(\left[ \begin{array}{l}x=1/3\\x=0\end{array} \right.\)
Vậy S={1/3;0}
c, (x+7)(3x-1)=49-x²
⇔(x+7)(3x-1)=(7-x)(7+x)
⇔(x+7)(3x-1)-(7-x)(x+7)=0
⇔(x+7)(3x-1-7+x)=0
⇔(x+7)(4x-8)=0
⇔(x+7).4(x-2)=0
⇔\(\left[ \begin{array}{l}x+7=0\\x-2=0\end{array} \right.\) <=>\(\left[ \begin{array}{l}x=-7\\x=2\end{array} \right.\)
Vậy S={-7;2}
d, x³-5x²+6x=0
⇔x(x²-5x+6)=0
⇔x(x²-3x-2x+6)=0
⇔x.[x(x-3)-2(x-3)]=0
⇔x(x-3)(x-2)=0
⇔x=0; x-3=0 hoặc x-2=0
⇔x=0; x=3 hoặc x=2
Vậy S={0;3;2}
e, $\frac{2x+1}{2x-1}-$ $\frac{2x-1}{2x+1}=$ $\frac{8}{4x^2-1}$ ⇔$\frac{(2x+1)(2x+1)-(2x-1)(2x-1)-8}{(2x+1)(2x-1)}=0$
⇔(2x+1)²-(2x-1)²-8=0
⇔(2x+1)²-(2x-1)²-8=0
⇔8(x-1)=0
⇔x-1=0
⇔x=1
Vậy S={1}
