Ta có \(\frac{a+b+c}{a+b-c}=\frac{a-b+c}{a-b-c}=\frac{a+b+c-\left(a-b+c\right)}{a+b-c-\left(a-b-c\right)}=\frac{2b}{2b}=1\)(dãy tỉ số bằng nhau)
Khi đó a + b + c = a + b - c
<=> c = - c
<=> 2 x c = 0
<=> c = 0(đpcm)
\(\frac{a+b+c}{a+b-c}=\frac{a-b+c}{a-b-c}\)
\(\left(a+b+c\right)\left(a-b-c\right)=\left(a-b+c\right)\left(a+b-c\right)\)
\(a^2+ab+ac-ab-b^2-bc-ac-bc-c^2=a^2+ab-ac-ab-b^2+bc+ac+cb-c^2\)
\(a^2-b^2-c^2-2bc=a^2-b^2-c^2+2bc\)
\(-2bc=2bc\)
mà \(b\ne0\)
thì \(-2bc;2bc\)trái dấu
vậy để \(-2bc=2bc\)thì \(c=0\)
\(< =>ĐPCM\)