Question

cho a, b, c là các số nguyên dương thỏa mãn a+b+c=2021

cmr:A ko phải một số nguyên, bt A=\(\dfrac{a}{2021-c}+\dfrac{b}{2021-a}+\dfrac{c}{2021-b}\)

 

Answer 1

\(\dfrac{a}{2021-c}+\dfrac{b}{2021-a}+\dfrac{c}{2021-b}\\ =\dfrac{a}{a+b+c-c}+\dfrac{b}{a+b+c-a}+\dfrac{c}{a+b+c-b}\\ =\dfrac{a}{a+b}+\dfrac{b}{b+c}+\dfrac{c}{c+a}\)

\(\dfrac{a}{a+b}+\dfrac{b}{b+c}+\dfrac{c}{c+a}>\dfrac{a}{a+b+c}+\dfrac{b}{a+b+c}+\dfrac{c}{a+b+c}=\dfrac{a+b+c}{a+b+c}=1\)

\(\dfrac{a}{a+b}+\dfrac{b}{b+c}+\dfrac{c}{c+a}< \dfrac{a+b}{a+b+c}+\dfrac{b+c}{a+b+c}+\dfrac{c+a}{a+b+c}=\dfrac{2\left(a+b+c\right)}{a+b+c}=2\)

Vì \(1< \dfrac{a}{a+b}+\dfrac{b}{b+c}+\dfrac{c}{c+a}< 2\Rightarrow A.ko.phải.số.nguyên\)