Question

[ỵyjfdfj]

Cho ba số a; b; c > 0 thỏa mãn: \(\dfrac{a+b-3c}{c}=\dfrac{b+c-3a}{a}=\dfrac{c+a-3b}{b}\)

Chứng minh rằng a = b =c.

Answer 1

[ILoveMath]

Áp dụng tính chất dãy tỉ số bằng nhau ta có:

\(\dfrac{a+b-3c}{c}=\dfrac{b+c-3a}{a}=\dfrac{c+a-3b}{b}=\dfrac{a+b-3c+b+c-3a+c+a-3b}{c+a+b}=\dfrac{-\left(a+b+c\right)}{a+b+c}=-1\)

\(\dfrac{a+b-3c}{c}=-1\Rightarrow a+b-3c=-c\Rightarrow a+b-2c=0\left(1\right)\)

\(\dfrac{b+c-3a}{a}=-1\Rightarrow b+c-3a=-a\Rightarrow b+c-2a=0\left(2\right)\)

\(\dfrac{c+a-3b}{b}=-1\Rightarrow a+c-3b=-b\Rightarrow a+c-2b=0\left(3\right)\)

Từ (1), (2) ta có:\(a+b-2c=b+c-2a\Rightarrow3a=3c\Rightarrow a=c\left(4\right)\)

Từ (1), (3) ta có:\(a+b-2c=a+c-2b\Rightarrow3b=3c\Rightarrow b=c\left(5\right)\)

Từ (4), (5)\(\Rightarrow a=b=c\)