Question

mình đg cần gấp câu d bài 4 nhé!

Answer 1

Bài 5: 

Ta có: \(\dfrac{158-x}{31}+\dfrac{185-x}{29}+\dfrac{208-x}{27}+\dfrac{227-x}{25}=10\)

\(\Leftrightarrow\dfrac{158-x}{31}+\dfrac{185-x}{29}+\dfrac{208-x}{27}+\dfrac{227-x}{25}-10=0\)

\(\Leftrightarrow\dfrac{158-x}{31}-1+\dfrac{185-x}{29}-2+\dfrac{208-x}{27}-3+\dfrac{227-x}{25}-4=0\)

\(\Leftrightarrow\dfrac{127-x}{31}+\dfrac{127-x}{29}+\dfrac{127-x}{27}+\dfrac{127-x}{25}=0\)

\(\Leftrightarrow\left(127-x\right)\left(\dfrac{1}{31}+\dfrac{1}{29}+\dfrac{1}{27}+\dfrac{1}{25}\right)=0\)

mà \(\dfrac{1}{31}+\dfrac{1}{29}+\dfrac{1}{27}+\dfrac{1}{25}>0\)

nên 127-x=0

hay x=127

Vậy: S={127}