\(\frac{a+b+c}{a+b-c}=\frac{a-b+c}{a-b-c}\Leftrightarrow\left(a+b+c\right).\left(a-b-c\right)=\left(a+b-c\right).\left(a-b+c\right)\)
\(\Leftrightarrow-c^2-2bc-b^2+a^2=-c^2+2bc-b^2+a^2\)
\(\Leftrightarrow-2bc=2bc\Rightarrow-bc=bc\Rightarrow\orbr{\begin{cases}b=0\\c=0\end{cases}}\)
=> đpcm
\(\frac{a+b+c}{a+b-c}=\frac{a-b+c}{a-b-c}\)
\(\Rightarrow\left(a+b+c\right)\left(a-b-c\right)=\left(a+b-c\right)\left(a-b+c\right)\)
\(\Rightarrow\left[a+\left(b+c\right)\right]\left[a-\left(b+c\right)\right]=\left[a+\left(b-c\right)\right]\left[a-\left(b-c\right)\right]\)
\(\Rightarrow a^2-\left(b+c\right)^2=a^2-\left(b-c\right)^2\)
\(\Rightarrow\left(b+c\right)^2=\left(b-c\right)^2\Rightarrow b^2+2bc+c^2=b^2-2bc+c^2\)
\(\Rightarrow2bc=-2bc\Rightarrow2bc+2bc=0\Rightarrow4bc=0\Rightarrow\orbr{\begin{cases}b=0\\c=0\end{cases}}\)
\(\frac{a+b+c}{a+b-c}=\frac{a-b+c}{a-b-c}\)
\(\Leftrightarrow\frac{a+b-c+2c}{a+b-c}=\frac{a-b-c+2c}{a-b-c}\)
\(\Leftrightarrow1+\frac{2c}{a+b-c}=1+\frac{2c}{a-b-c}\)
\(\Leftrightarrow\frac{c}{a+b-c}=\frac{c}{a-b-c}\)
*Nếu \(c=0\)thì đẳng thức trên luôn đúng
*Nếu \(c\ne0\)thì \(a+b-c=a-b-c\)
\(\Leftrightarrow2b=0\Leftrightarrow b=0\)
Vậy b = 0 hoặc c = 0 (đpcm)
