Giải các phương trình sau: $\frac{x+1}{99}$ + $\frac{x+3}{97}$ + $\frac{x+5}{95}$ = $\frac{x+7}{93}$ + $\frac{x+9}{91}$ + $\frac{x+11}{89}$

$\dfrac{x+1}{99}+\dfrac{x+3}{97}+\dfrac{x+5}{95}=\dfrac{x+7}{93}+\dfrac{x+9}{91}+\dfrac{x+11}{89}$ 

$\dfrac{x+1}{99}+1+\dfrac{x+3}{97}+1+\dfrac{x+5}{95}+1=\dfrac{x+7}{93}+1+\dfrac{x+9}{91}+1+\dfrac{x+11}{89}+1$ 

$\dfrac{x+100}{99}+\dfrac{x+100}{97}+\dfrac{x+100}{95}=\dfrac{x+100}{93}+\dfrac{x+100}{91}+\dfrac{x+100}{89}$ 

$\dfrac{x+100}{99}+\dfrac{x+100}{97}+\dfrac{x+100}{95}-\dfrac{x+100}{93}-\dfrac{x+100}{91}-\dfrac{x+100}{89}=0$ 

$(x+100)(\dfrac{1}{99}+\dfrac{1}{97}+\dfrac{1}{95}-\dfrac{1}{93}-\dfrac{1}{91}-\dfrac{1}{89})=0$ 

Vì: $\dfrac{1}{99}+\dfrac{1}{97}+\dfrac{1}{95}-\dfrac{1}{93}-\dfrac{1}{91}-\dfrac{1}{89} \neq 0$ 

$⇔x+100=0$

$⇔x=-100$

Vậy phương trình có nghiệm `S={-100}`

Đáp án:

Vậy, phương trình đã cho có tập nghiệm là `S={-100}.`

Giải thích các bước giải:

`(x+1)/99+(x+3)/97+(x+5)/95=(x+7)/93+(x+9)/91+(x+11)/89`

`<=>(x+1)/99+(x+3)/97+(x+5)/95-(x+7)/93-(x+9)/91-(x+11)/89=0`

`<=>((x+1)/99+1)+((x+3)/97+1)+((x+5)/95+1)-((x+7)/93+1)-((x+9)/91+1)-((x+11)/89+1)=0`

`<=>((x+1)/99+1/1)+((x+3)/97+1/1)+((x+5)/95+1/1)-((x+7)/93+1/1)-((x+9)/91+1/1)-((x+11)/89+1/1)=0`

`<=>((x+1)/99+99/99)+((x+3)/97+97/97)+((x+5)/95+95/95)-((x+7)/93+93/93)-((x+9)/91+91/91)-((x+11)/89+89/89)=0`

`<=>((x+1+99)/99)+((x+3+97)/97)+((x+5+95)/95)-((x+7+93)/93)-((x+9+91)/91)-((x+11+89)/89)=0`

`<=>(x+100)/99+(x+100)/97+(x+100)/95-(x+100)/93-(x+100)/91-(x+100)/89=0`

`<=>(x+100).(1/99+1/97+1/95-1/93-1/91-1/89)=0`

`<=>x+100=0`

`<=>x=0-100`

`<=>x=-100`

Vậy, phương trình đã cho có tập nghiệm là `S={-100}.`