Đáp án:

`↓↓↓`

Giải thích các bước giải:

`1)`

`(x-1)/(999)+(x-2)/(499)+(x-4)/(332)=6`

`⇔(x-1)/(999)+(x-2)/(499)+(x-4)/(332)-6=0`

`⇔((x-1)/(999)-1)+((x-2)/(499)-2)+((x-4)/(332)-3)=0`

`⇔(x-1000)/(999)+(x-1000)/(499)+(x-1000)/(332)=0`

`⇔(x-1000).((1)/(999)+(1)/(499)+(1)/(332))=0`

Do : `(1)/(999)+(1)/(499)+(1)/(332)` `\ne0` 

Nên `x-1000=0` là nghiệm của phương trình `⇔x=1000`

Vậy phương trình trên có nghiệm là `S={1000}`

`2)`

`(1)/(x^2+9x+20)+(1)/(x^2+11x+30)+(1)/(x^2+13x+42)=1/(18)`

`⇔(1)/((x+4)(x+5))+(1)/((x+5)(x+6))+(1)/((x+6)(x+7))=1/(18)`

`ĐKXĐ:{(x\ne-4),(x\ne-5),(x\ne-6),(x\ne-7):}`

`⇔(1)/(x+4)-(1)/(x+5)+(1)/(x+5)-(1)/(x+6)+(1)/(x+6)-(1)/(x+7)=1/(18)`

`⇔(1)/(x+4)-(1)/(x+7)=1/(18)`

`⇔(x+7-x-4)/((x+4)(x+7))=1/(18)`

`⇔(3)/((x+4)(x+7))=1/(18)`

`⇔(x+4)(x+7)=3.18`

`⇔x^2+7x+4x+28=54`

`⇔x^2+11x-26=0`

`⇔x^2-2x+13x-26=0`

`⇔x(x-2)+13(x-2)=0`

`⇔(x-2)(x+13)=0`

`⇔`$\left[\begin{matrix} x-2=0\\ x+13=0\end{matrix}\right.$

`⇔`$\left[\begin{matrix} x=2(tmđk)\\ x=-13(tmđk)\end{matrix}\right.$

Vậy phương trình trên có nghiệm là `S={2;-13}`

Giải thích các bước giải:

1.   $\dfrac{x-1}{999}$ + $\dfrac{x-2}{499}$ + $\dfrac{x-4}{332}$ = 6

⇔  $\dfrac{x-1}{999}$ -1 + $\dfrac{x-2}{499}$ -2 + $\dfrac{x-4}{332}$ -3  = 0 

⇔  $\dfrac{x-1000}{999}$ +$\dfrac{x-1000}{499}$ + $\dfrac{x-1000}{332}$ = 0

⇔  (x-100).($\dfrac{1}{999}$ + $\dfrac{1}{499}$ + $\dfrac{1}{332}$) = 0

⇔            x - 1000 = 0  ( do ($\dfrac{1}{999}$ + $\dfrac{1}{499}$ + $\dfrac{1}{332}$) $\neq$ 0)

⇔            x = 1000

Vậy x = 1000

2. Điều kiện: x khác -4,-5,-6,-7

$\dfrac{1}{x² +9x +20}$ + $\dfrac{1}{x² +11x +30}$ + $\dfrac{1}{x² +13x +42}$ = $\dfrac{1}{18}$

⇔  $\dfrac{1}{x² +4x +5x +20}$+$\dfrac{1}{x² +5x +6x +30}$+$\dfrac{1}{x² +6x +7x +42}$ = $\dfrac{1}{18}$

⇔ $\frac{1}{x.(x+4) +5.(x+4)}$ +$\dfrac{1}{x.(x+5) +6.(x+5)}$+$\dfrac{1}{x.(x+6) +7.(x+6)}$ = $\dfrac{1}{18}$

⇔ $\dfrac{1}{(x+4).(x+5)}$ + $\dfrac{1}{(x+5).(x+6)}$ + $\dfrac{1}{(x+6).(x+7)}$ = $\dfrac{1}{18}$

⇔ $\dfrac{1}{x+4}$ -$\dfrac{1}{x+5}$ +$\dfrac{1}{x+5}$ -$\dfrac{1}{x+6}$ +$\dfrac{1}{x+6}$ -$\dfrac{1}{x+7}$ = $\dfrac{1}{18}$

⇔ $\dfrac{1}{x+4}$ -$\dfrac{1}{x+7}$ = $\dfrac{1}{18}$

⇔ $\dfrac{x+7 -x-4}{(x+4).(x+7)}$ = $\dfrac{1}{18}$

⇔  (x+4).(x+7) = 3.18

⇔ $\text{x² +11x +28 -54 = 0}$

⇔ $\text{x² -2x +13x -26 = 0}$

⇔ $\text{x.(x-2) + 13.(x-2) = 0}$

⇔ $\text{(x-2).(x+13) = 0}$

⇔ \(\left[ \begin{array}{l}x-2 = 0\\x+13 = 0\end{array} \right.\) 

⇔ \(\left[ \begin{array}{l}x=2 ( tmdk)\\x=-13 (tmdk)\end{array} \right.\) 

Vậy x∈{-13; 2}