Bài 2:
\(\dfrac{a+b+c}{a+b-c}=\dfrac{a-b+c}{a-b-c}\)
=>\(\left(a+b+c\right)\left(a-b-c\right)=\left(a+b-c\right)\left(a-b+c\right)\)
=>\(\left[a+\left(b+c\right)\right]\left[a-\left(b+c\right)\right]=\left[a+\left(b-c\right)\right]\left[a-\left(b-c\right)\right]\)
=>\(a^2-\left(b+c\right)^2=a^2-\left(b-c\right)^2\)
=>\(\left(b+c\right)^2=\left(b-c\right)^2\)
=>\(b^2+2bc+c^2=b^2-2bc+c^2\)
=>4bc=0
=>bc=0
mà b<>0
nên c=0
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