Đáp án + Giải thích các bước giải:
`(4x+1)(12x-1)(3x+2)(x+1)-4=0`
`<=>(4x+1)(3x+2)(12x-1)(x+1)-4=0`
`<=>(12x^2+11x+2)(12x^2+11x-1)-4=0`
Đặt `12x^2+11x+2=t(t>=0)`
`<=>t(t+3)-4=0`
`<=>t^2+3t-4=0`
`<=>t^2-t+4t-4=0`
`<=>t(t-1)+4(t-1)=0`
`<=>(t-1)(t+4)=0`
`<=>`\(\left[ \begin{array}{l}t=1(tm)\\t=-4(ktm)\end{array} \right.\)
Do đó : `12x^2+11x+2=1<=>12x^2+11x+1=0`
`<=>12(x^2+11/12x+1/12)=0`
`<=>12[x^2+2*x*11/24+(11/24)^2]-73/48=0`
`<=>12(x+11/24)^2=73/48`
`<=>(x+11/24)^2=73/576`
`<=>` `(x+11/24)^2=pm(sqrt73)/24`
TH1 : `x + 11/24=(sqrt73)/24=>x=(sqrt73)/24-11/24=(sqrt73-11)/24`
TH2 : `x + 11/24=-(sqrt73)/24<=>x=-(sqrt73)/24-11/24=-(sqrt73-11)/24`
Vậy `S={(sqrt73-11)/24,-(sqrt73-11)/24}`
