`ĐKXĐ: x ne -1, -2, -3, -4, -5` 

`<=> 1/(x^2 + x + 2x + 2) + 1/(x^2+2x+ 3x + 6) + 1/(x^2 + 3x+ 4x + 12) + 1/(x^2 + 4x + 5x + 20) = 4/45`

`<=> 1/((x+1)(x+2)) + 1/((x+2)(x+3)) + 1/((x+3)(x+4)) + 1/((x+4)(x+5)) = 4/45`

`<=> 1/(x+1) - 1/(x+2) + 1/(x+2) - 1/(x+3) + 1/(x +3)- 1/(x+4) + 1/(x+4) - 1/(x+5) = 4/45`

`<=> 1/(x+1) -1/(x+5) = 4/45`

`<=>4/((x+1)(x+5)) = 4/45`

`<=> x^2 + 6x + 5 = 45`

`<=> x^2 + 6x  = 40`

`<=> x^2 + 6x + 9 = 49`

`<=> (x+3)^2 = 49`

`<=> x+3 = ±7`

`TH_1: x+3 = 7`

`<=> x = 7 - 3 = 4`

`TH_2: x+3 = -7`

`<=> x = -7-3 = -10`

Vậy khi `x = -10, 4` thì PT thỏa mãn.

Đáp án:

 `h)S={4;-10}`

Giải thích các bước giải:

 `h)1/(x^2+3x+2)+1/(x^2+5x+6)+1/(x^2+7x+12)+1/(x^2+9x+20)=4/45`(ĐKXĐ: `x\ne-1;-2;-3;-4;-5)`

`<=>1/(x^2+x+2x+2)+1/(x^2+2x+3x+6)+1/(x^2+3x+4x+12)+1/(x^2+4x+5x+20)=4/45`

`<=>1/(x(x+1)+2(x+1))+1/(x(x+2)+3(x+2))+1/(x(x+3)+4(x+3))+1/(x(x+4)+5(x+4))=4/45`

`<=>1/((x+1)(x+2))+1/((x+2)(x+3))+1/((x+3)(x+4))+1/((x+4)(x+5))=4/45`

`<=>1/(x+1)-1/(x+2)+1/(x+2)-1/(x+3)+1/(x+3)-1/(x+4)+1/(x+4)-1/(x+5)=4/45`

`<=>1/(x+1)-1/(x+5)=4/45`

`<=>4/((x+1)(x+5))=4/45`

`<=>(x+1)(x+5)=45`

`<=>x^2+6x+5=45`

`<=>x^2+6x-40=0`

`<=>x^2+6x+9-49=0`

`<=>(x+3-7)(x+3+7)=0`

`<=>(x-4)(x+10)=0`

`<=>`\(\left[ \begin{array}{l}x-4=0\\x+10=0\end{array} \right.\)

`<=>`\(\left[ \begin{array}{l}x=4(t/m)\\x=-10(t/m)\end{array} \right.\)

`@nguyen``nam500#hoidap247`