Câu 5:

\(\dfrac{a+b-c}{c}=\dfrac{a-b+c}{b}=\dfrac{-a+b+c}{a}\)

=>\(\dfrac{a+b}{c}-1=\dfrac{a+c}{b}-1=\dfrac{b+c}{a}-1\)=>\(\dfrac{a+b}{c}=\dfrac{a+c}{b}=\dfrac{b+c}{a}\)

=>\(\dfrac{a+b}{c}=\dfrac{a+c}{b}=\dfrac{b+c}{a}=\dfrac{a+b+a+c+b+c}{a+b+c}=\dfrac{2\left(a+b+c\right)}{a+b+c}=2\)

=>\(\dfrac{\left(a+b\right)\left(b+c\right)\left(c+a\right)}{abc}=2.2.2=8\)

- Đây là đề giữa kì mà?

- Giữa kì rồi mà ?

- Giữa kì rồi mà ?