đặt:\(\dfrac{a}{c}=\dfrac{c}{b}=k\)⇒ c= bk; a= ck= bk2
ta có: \(\dfrac{a^2+c^2}{b^2+c^2}=\dfrac{\left(bk^2\right)^2+\left(bk\right)^2}{b^2+\left(bk\right)^2}=\dfrac{b^2k^2k^2+b^2k^2}{b^2+b^2k^2}\)
=\(\dfrac{\left(b^2k^2+b^2\right)k^2}{b^2\left(k^2+1\right)}=\dfrac{\left[b^2\left(k^2+1\right)\right]k^2}{b^2\left(k^2+1\right)}=k^2\)
\(k^2=\dfrac{a}{c}\times\dfrac{a}{c}=\dfrac{c}{b}\times\dfrac{c}{b}=\dfrac{a}{c}\times\dfrac{c}{b}=\dfrac{a}{b}\)
Vậy\(\dfrac{a^2+c^2}{b^2+c^2}=\dfrac{a}{b}\)
có gì sai gì bạn thông cảm nhé