`a)` `x^3-6x^2+10x-8=0`
`<=>x^3-4x^2-2x^2+8x+2x-8=0`
`<=>x^2(x-4)-2x(x-4)+2(x-4)=0`
`<=>(x^2-2x+2)(x-4)=0`
Xét `x^2-2x+2`
`=x^2-2x+1+1`
`=(x-1)^2+1\geq1>0` `text{( loại )}`
Xét: `x-4=0`
`<=>x=4`
Vậy `S={4}`
`b)` `frac{3}{2x+1}-frac{5}{1-2x}=frac{9-4x}{4x^2-1}` ĐKXĐ: `x\ne±1/2`
`<=>frac{3}{2x+1}+frac{5}{2x-1}=frac{9-4x}{(2x-1)(2x+1)}`
`<=>frac{3(2x-1)}{(2x+1)(2x-1)}+frac{5(2x+1)}{(2x+1)(2x-1)}=frac{9-4x}{(2x-1)(2x+1)}`
`=>3(2x-1)+5(2x+1)=9-4x`
`<=>6x-3+10x+5=9-4x`
`<=>16x+2=9-4x`
`<=>16x+4x=9-2`
`<=>20x=7`
`<=>x=7/20` `text{( thoả mãn điều kiện )}`
Vậy `S={7/20}`
