Đáp án:  `S={50}`

Giải thích các bước giải:

        `(29-x)/21 + (27-x)/23 + (25-x)/25 + (23-x)/27+ (21-x)/29=-5`

$\text{ ⇔ ( $\dfrac{29-x}{21}$+1) + ( $\dfrac{27-x}{23}$+1) + ( $\dfrac{25-x}{25}$+1 )+ ( $\dfrac{23-x}{27}$+1) + ( $\dfrac{21-x}{29}$ +1) =0}$ 

$\text{ ⇔$\dfrac{29-x+21}{21}$+ $\dfrac{27-x+23}{23}$ + $\dfrac{25-x+25}{25}$ + $\dfrac{23-x+27}{27}$ + $\dfrac{21-x+29}{29}$ =0 }$

$\text{ ⇔$\dfrac{50-x}{21}$ + $\dfrac{50-x}{23}$ + $\dfrac{50-x}{25}$ + $\dfrac{50-x}{27}$ + $\dfrac{50-x}{29}$=0 }$   

$\text{ ⇔(50-x)( $\dfrac{1}{21}$ + $\dfrac{1}{23}$ + $\dfrac{1}{25}$ + $\dfrac{1}{27}$ + $\dfrac{1}{29}$ =0 }$

Vì `1/21+1/23+1/25+1/27+1/29≠0`

`⇔50-x=0`

`⇔x=50`

Vậy, `S={50}`

`#` `ccUh`