Đáp án+Giải thích các bước giải:
$\\\text{Từ $\dfrac{ab}{bc} =\dfrac{b}{c}=>\dfrac{ab}{b}=\dfrac{bc}{c} (1)$}\\=>\dfrac{10a+b}{b}=\dfrac{10b+c}{c}\\=>\dfrac{10a}{b}+1=\dfrac{10b}{c}+1\\=>\dfrac{10a}{b}=\dfrac{10b}{c}=\dfrac{a}{b}=\dfrac{b}{c}\\=>a.c=b^2\\+)\dfrac{a^2+b^2}{b^2+c^2}=\dfrac{a^2+ a.c}{c^2+a.c}=\dfrac{a.(a+c)}{c.(c+a)}=\dfrac{a}{c}$