`a)(x-2)/3-(x-4)/5=(x-5)/6`
`⇔[10(x-2)]/30-[6(x-4)]/30=[5(x-5)]/30`
`⇔10(x-2)-6(x-4)=5(x-5)`
`⇔10x-20-6x+24=5x-25`
`⇔(10x-6x)+(-20+24)=5x-25`
`⇔4x+4=5x-25`
`⇔4x-5x=-25-4`
`⇔-x=-29`
`⇔x=29`
Vậy `S={29}`
`b)(3x-2)²=(2-3x)(4x+5)`
`⇔(3x-2)²-(2-3x)(4x+5)=0`
`⇔(3x-2)²+(3x-2)(4x+5)=0`
`⇔(3x-2)(3x-2+4x+5)=0`
`⇔(3x-2)(7x+3)=0`
`⇔`$\left[\begin{matrix} 3x-2=0\\ 7x+3=0\end{matrix}\right.$
`⇔`$\left[\begin{matrix} 3x=2\\ 7x=-3\end{matrix}\right.$
`⇔`$\left[\begin{matrix} x=\dfrac{2}{3}\\ x=-\dfrac{3}{7}\end{matrix}\right.$
Vậy `S={2/3;-3/7}`
`c)(x+3)/(x-3)+36/(9-x²)=(x-3)/(x+3)(ĐKXĐ:x`$\neq$ `±3)`
`⇔(x+3)/(x-3)-36/(x²-9)=(x-3)/(x+3)`
`⇔[(x+3)²]/(x²-9)-36/(x²-9)=[(x-3)²]/(x²-9)`
`⇒(x+3)²-36=(x-3)²`
`⇔x²+6x+9-36=x²-6x+9`
`⇔x²+6x-x²+6x=9-9+36`
`⇔12x=36`
`⇔x=36:12`
`⇔x=3(Ko` `TM` `ĐKXĐ)`
Vậy `S=∅`
`d)3x²-4x=5(4-3x)`
`⇔x(3x-4)=-5(3x-4)`
`⇔x(3x-4)+5(3x-4)=0`
`⇔(x+5)(3x-4)=0`
`⇔`$\left[\begin{matrix} x+5=0\\ 3x-4=0\end{matrix}\right.$
`⇔`$\left[\begin{matrix} x=-5\\ x=\dfrac{4}{3}\end{matrix}\right.$
Vậy `S={-5;4/3}`
`e)2x²-3x-9=0`
`⇔2x²-6x+3x-9=0`
`⇔2x(x-3)+3(x-3)=0`
`⇔(x-3)(2x+3)=0`
`⇔`$\left[\begin{matrix} x-3=0\\ 2x+3=0\end{matrix}\right.$
`⇔`$\left[\begin{matrix} x=3\\ x=-\dfrac{3}{2}\end{matrix}\right.$
Vậy `S={3;-3/2}`
`f)(x+1)/(x-1)-(4x³)/(x²-1)=(x-1)/(x+1)(ĐKXĐ:x`$\neq$ `±1)`
`⇔[(x+1)²]/(x²-1)-(4x³)/(x²-1)=[(x-1)²]/(x²-1)`
`⇒(x+1)²-4x³=(x-1)²`
`⇔x²+2x+1-4x³=x²-2x+1`
`⇔x²+2x+1-4x³-x²+2x-1=0`
`⇔-4x³+(x²-x²)+(2x+2x)+(1-1)=0`
`⇔-4x³+4x=0`
`⇔-4x(x²-1)=0`
`⇔-4x(x+1)(x-1)=0`
`⇔`$\left[\begin{matrix} -4x=0\\x+1=0\\ x-1=0\end{matrix}\right.$
`⇔`$\left[\begin{matrix} x=0(TM ĐKXĐ)\\x=-1(Ko TM ĐKXĐ)\\ x=1(Ko TM ĐKXĐ)\end{matrix}\right.$
Vậy `S={0}`
