Đáp án:
`S=\{\frac{3\sqrt5-5}{2};\frac{-3\sqrt5-5}{2}\}`
Giải thích các bước giải:
Ta có:
`x^2+3x+2=x^2+x+2x+2=x(x+1)+2(x+1)=(x+1)(x+2)`
`x^2+5x+6=x^2+2x+3x+6=x(x+2)+3(x+2)=(x+2)(x+3)`
`x^2+7x+12=x^2+3x+4x+12=x(x+3)+4(x+3)=(x+3)(x+4)`
`\to ĐKXĐ: x\ne-1;x\ne-2;x\ne-3;x\ne-4`
`1/(x^2+3x+2)+1/(x^2+5x+6)+1/(x^2+7x+12)=1/3`
`⇔1/((x+1)(x+2))+1/((x+2)(x+3))+1/((x+3)(x+4))=1/3`
`⇔1/(x+1)-1/(x+2)+1/(x+2)-1/(x+3)+1/(x+3)-1/(x+4)=1/3`
`⇔1/(x+1)-1/(x+4)=1/3`
`⇔(x+4-(x+1))/((x+1)(x+4))=1/3`
`⇔(x+4-x-1)/((x+1)(x+4))=1/3`
`⇔3/((x+1)(x+4))=1/3`
`⇔9=(x+1)(x+4)`
`⇔x^2+5x+4=9`
`⇔x^2+5x-5=0`
`⇔x^2+2.(5)/2x+25/4-45/4=0`
`⇔(x+5/2)^2-45/4=0`
`⇔(x+5/2)^2=45/4`
`⇔(x+5/2)^2=((3\sqrt5)/2)^2`
`⇔|x+5/2|=(3\sqrt5)/2`
\(⇔\left[ \begin{array}{l}x=\dfrac{3\sqrt5-5}{2}(TM)\\x=\dfrac{-3\sqrt5-5}{2}(TM)\end{array} \right.\)
Vậy `S=\{\frac{3\sqrt5-5}{2};\frac{-3\sqrt5-5}{2}\}`
