Question

Giải phương trình

1. \(\dfrac{1909-x}{91}+\dfrac{1907-x}{93}+\dfrac{1905-x}{95}+\dfrac{1903-x}{91}+4=0\)

Answer 1

\(\dfrac{1909-x}{91}+\dfrac{1907-x}{93}+\dfrac{1905-x}{95}+\dfrac{1903}{97}+4=0\) ( Sửa đề)

⇔ \(\dfrac{1909-x}{91}+1+\dfrac{1907-x}{93}+1+\dfrac{1905-x}{95}+1+\dfrac{1903}{97}+1=0\) ⇔ \(\dfrac{2000-x}{91}+\dfrac{2000-x}{93}+\dfrac{2000-x}{95}+\dfrac{2000-x}{97}=0\)

⇔ \(\left(2000-x\right)\left(\dfrac{1}{91}+\dfrac{1}{93}+\dfrac{1}{95}+\dfrac{1}{97}\right)=0\)

Do : \(\dfrac{1}{91}+\dfrac{1}{93}+\dfrac{1}{95}+\dfrac{1}{97}>0\)

\(\text{⇔}2000-x=0\)

\(\text{⇔}x=2000\)

Vậy ,....