Question

Giải phương trình:

a.\(\dfrac{10-x}{100}+\dfrac{20-x}{110}+\dfrac{30-x}{120}=3\)

Answer 1

-90

Answer 2

\(\dfrac{10-x}{100}\) + \(\dfrac{20-x}{110}\)+\(\dfrac{30-x}{120}\)=3

<=> \(\dfrac{10-x}{100}\)-1+\(\dfrac{20-x}{110}\)-1+\(\dfrac{30-x}{120}\)-1 = 0

<=> \(\dfrac{-x-90}{100}\)+\(\dfrac{-x-90}{110}\)+\(\dfrac{-x-90}{120}\)=0

<=> (-x-90) ( \(\dfrac{1}{100}\)+\(\dfrac{1}{110}\)+\(\dfrac{1}{120}\))=0

<=> (-x-90) = 0 ( do 1/100 +1/110+1/120 khác 0)

<=> -x-90 = 0

<=> -x = 90

<=> x =-90

Vậy nghiệm của pt là x=-90