Question

cho 3 số a,b,c khác 0 thỏa mãn : \(\dfrac{2022a+b+c}{a}\) = \(\dfrac{a+2022b+c}{b}\) = \(\dfrac{a+b+2022c}{c}\) . tính giá trị của biểu thức P = \(\dfrac{a+b}{c}\) = \(\dfrac{b+c}{a}\) = \(\dfrac{a+c}{b}\)

Answer 1

Lời giải:

$\frac{2022a+b+c}{a}=\frac{a+2022b+c}{b}=\frac{a+b+2022c}{c}$

$=2021+\frac{a+b+c}{a}=2021+\frac{a+b+c}{b}=2021+\frac{a+b+c}{c}$

$\Rightarrow \frac{a+b+c}{a}=\frac{a+b+c}{b}=\frac{a+b+c}{c}$

$\Rightarrow a+b+c=0$ hoặc $\frac{1}{a}=\frac{1}{b}=\frac{1}{c}$

$\Rightarrow a+b+c=0$ hoặc $a=b=c$

Nếu $a+b+c=0$ thì:

$P=\frac{a+b}{c}+\frac{b+c}{a}+\frac{a+c}{b}=\frac{(-c)}{c}+\frac{(-b)}{b}+\frac{(-a)}{a}=-1+(-1)+(-1)=-3$
Nếu $a=b=c$ thì:

$P=\frac{c+c}{c}+\frac{a+a}{a}+\frac{b+b}{b}=2+2+2=6$