Nhầm....(-2)

Vì \(\frac{a}{c}=\frac{c}{b}=>ab=c^2\)

a)Ta có: \(\frac{a^2+c^2}{b^2+c^2}=\frac{a.a+ab}{b.b+ab}=\frac{a.\left(a+b\right)}{b.\left(a+b\right)}=\frac{a}{b}\)

=>\(\frac{a^2+c^2}{b^2+c^2}=\frac{a}{b}\)

=>ĐPCM

b)Ta có: \(\frac{b^2-a^2}{a^2+c^2}=\frac{\left(b-a\right).\left(b+a\right)}{a.a+ab}=\frac{\left(b-a\right).\left(b+a\right)}{a.\left(b+a\right)}=\frac{b-a}{a}\)

=>\(\frac{b^2-a^2}{a^2+c^2}=\frac{b-a}{a}\)

=>ĐPCM