Đáp án + Giải thích các bước giải:

`(1)/(x^{2}+3x+2)+(1)/(x^{2}+5x+6)+(1)/(x^{2}+7x+12)=(1)/(3)` `(ĐKXĐ:x\ne{-1;-2;-3;-4})`

`<=>(1)/((x^{2}+x)+(2x+2))+(1)/((x^{2}+2x)+(3x+6))+(1)/((x^{2}+3x)+(4x+12))=(1)/(3)`

`<=>(1)/(x(x+1)+2(x+1))+(1)/(x(x+2)+3(x+2))+(1)/(x(x+3)+4(x+3))=(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)`

`<=>(3)/((x+1)(x+4))=(1)/(3)`

`<=>(x+1)(x+4)=9`

`<=>x^{2}+x+4x+4=9`

`<=>x^{2}+5x-5=0`

`<=>(x^{2}+2.x.(5)/(2)+(25)/(4))-(45)/(4)=0`

`<=>(x+(5)/(2))^{2}=(45)/(4)`

`<=>x+(5)/(2)=±(3\sqrt{5})/(2)`

`<=>x=(-5±3\sqrt{5})/(2)`

Vậy phương trình có tập nghiệm là : `S={(-5±3\sqrt{5})/(2)}`

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}\}`