\(đat:\frac{a}{b}=\frac{c}{d}=k\Rightarrow\left\{{}\begin{matrix}a=bk\\c=dk\end{matrix}\right.\)
\(a,\frac{a^2-b^2}{ab}=\frac{b^2k^2-b^2}{bkb}=\frac{b^2\left(k^2-1\right)}{b^2k}=\frac{k^2-1}{k};\frac{c^2-d^2}{cd}=\frac{d^2\left(k^2-1\right)}{d^2k}=\frac{k^2-1}{k}\Rightarrow\frac{a^2-b^2}{ab}=\frac{c^2-d^2}{cd}\) \(b,\frac{\left(a+b\right)^2}{a^2+b^2}=\frac{\left[b\left(k+1\right)\right]^2}{b^2k^2+b^2}=\frac{b^2\left(k+1\right)^2}{b^2\left(k^2+1\right)}=\frac{\left(k+1\right)^2}{\left(k^2+1\right)};\frac{\left(c+d\right)^2}{c^2+d^2}=\frac{\left[d\left(k+1\right)\right]^2}{d^2k^2+d^2}=\frac{d^2\left(k+1\right)^2}{d^2\left(k^2+1\right)}=\frac{\left(k+1\right)^2}{k^2+1}\Rightarrow\frac{\left(a+b\right)^2}{a^2+b^2}=\frac{\left(c+d\right)^2}{c^2+d^2}\) \(c,\frac{a}{a+b}=\frac{bk}{bk+b}=\frac{bk}{b\left(k+1\right)}=\frac{k}{k+1};\frac{c}{c+d}=\frac{dk}{dk+d}=\frac{dk}{d\left(k+1\right)}=\frac{k}{k+1}\Rightarrow\frac{a}{a+b}=\frac{c}{c+d}\)
