Bài làm:
Ta có: \(\frac{a}{b}=\frac{b}{c}\) => \(b^2=ac\)
Thay vào ta được: \(\frac{a^2+b^2}{b^2+c^2}=\frac{a^2+ac}{ac+c^2}=\frac{a\left(a+c\right)}{c\left(a+c\right)}=\frac{a}{c}\)
Vậy \(\frac{a}{c}=\frac{a^2+b^2}{b^2+c^2}\)
Đặt \(\frac{a}{b}=\frac{b}{c}=k\Rightarrow\hept{\begin{cases}a=kb\\b=kc\end{cases}}\Rightarrow a=k^2c\)
\(\frac{a^2+b^2}{b^2+c^2}=\frac{\left(k^2c\right)^2+\left(kc\right)^2}{\left(kb\right)^2+c^2}=\frac{k^4b^c+k^2c^2}{k^2b^2+c^2}=\frac{k^2\left(k^2b^2+c^2\right)}{k^2b^2+c^2}=k^2=\frac{a}{c}\)( đpcm )
Ý sai -..- sửa lại nghen :v
\(\frac{a^2+b^2}{b^2+c^2}=\frac{\left(k^2c\right)^2+\left(kc\right)^2}{\left(kc\right)^2+c^2}=\frac{k^4c^2+k^2c^2}{k^2c^2+c^2}=\frac{k^2\left(k^2c^2+c^2\right)}{k^2c^2+c^2}=k^2=\frac{a}{c}\)( đpcm )
Xin lỗi bạn vì sự nhầm lẫn : ((
