$\frac{x-1}{99}$ + $\frac{x-2}{49}$ + $\frac{x-7}{31}$ + $\frac{x-8}{23}$ =10

Đáp án + Giải thích các bước giải:

 `(x-1)/99 + (x-2)/49 +(x-7)/31 + (x-8)/23=10`

`⇔ (x-1)/99+ (x-2)/49 +(x-7)/31 + (x-8)/23-10=0`

`⇔ (x-1)/99 -1 + (x-2)/49 -2 +(x-7)/31 -3 + (x-8)/23 -4 =0`

`⇔ (x-1)/99 -99/99 + (x-2)/49 -98/49+(x-7)/31 -93/31+ (x-8)/23 - 92/23=0`

`⇔ (x-1-99)/99+ (x-2-98)/49 +(x-7-93)/31 + (x-8-92)/23 =0`

`⇔ (x-100)/99+ (x-100)/49 +(x-100)/31 + (x-100)/23=0`

`⇔ (x-100).(1/99+1/49+1/31+1/23)=0`

`⇔ x-100=0` vì `1/99+1/49+1/31+1/23\ne0`

`⇔ x=0+100`

`⇔ x=100`

Vậy `S={100}`