Question

giải pt

\(\dfrac{201-x}{99}+\dfrac{203-x}{97}=\dfrac{205-x}{95}+3=0\)

Answer 1

Sửa đề: \(\dfrac{201-x}{99}+\dfrac{203-x}{97}+\dfrac{205-x}{95}+3=0\)

Ta có: \(\dfrac{201-x}{99}+\dfrac{203-x}{97}+\dfrac{205-x}{95}+3=0\)

\(\Leftrightarrow\dfrac{201-x}{99}+1+\dfrac{203-x}{97}+1+\dfrac{205-x}{95}+1=0\)

\(\Leftrightarrow\dfrac{300-x}{99}+\dfrac{300-x}{97}+\dfrac{300-x}{95}=0\)

\(\Leftrightarrow\left(300-x\right)\left(\dfrac{1}{99}+\dfrac{1}{97}+\dfrac{1}{95}\right)=0\)

mà \(\dfrac{1}{99}+\dfrac{1}{97}+\dfrac{1}{95}>0\)

nên 300-x=0

hay x=300

Vậy:S={300}