Do \(\dfrac{a}{c}=\dfrac{c}{b}\)
\(\Rightarrow ab=c^2\)
Ta có: \(\dfrac{b^2-a^2}{a^2+c^2}=\dfrac{b^2-a^2}{a^2+ab}=\dfrac{\left(b^2-ab\right)+\left(ab-a^2\right)}{a\left(a+b\right)}=\dfrac{b\left(b-a\right)+a\left(b-a\right)}{a\left(a+b\right)}=\dfrac{\left(b+a\right)\left(b-a\right)}{a\left(a+b\right)}=\dfrac{b-a}{a}\)
\(\dfrac{a}{c}=\dfrac{c}{b}\\ \Rightarrow ab=c^2\)
\(\dfrac{b^2-a^2}{a^2+c^2}=\dfrac{\left(b-a\right)\left(b+a\right)}{a^2+ab}=\dfrac{\left(b-a\right)\left(a+b\right)}{a\left(a+b\right)}=\dfrac{b-a}{a}\)