Giải thích các bước giải:

Ta có:

$\dfrac{29-x}{21}+\dfrac{27-x}{23}+\dfrac{25-x}{25}+\dfrac{23-x}{27}+\dfrac{21-x}{29}=-5$ 

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

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

$\to (50-x)\cdot (\dfrac1{21}+\dfrac1{23}+\dfrac1{25}+\dfrac1{27}+\dfrac1{29})=0$

$\to 50-x=0$

$\to x=50$