\(\frac{a}{b}=\frac{c}{d}\Rightarrow\frac{a}{c}=\frac{b}{d}\Rightarrow\left(\frac{a}{c}\right)^2=\left(\frac{b}{d}\right)^2=\frac{\left(a-b\right)^2}{\left(c-d\right)^2}\)
                                         \(\Rightarrow\frac{a^2}{c^2}=\frac{b^2}{d^2}=\frac{\left(a-b\right)^2}{\left(c-d\right)^2}\)
                                          \(\Rightarrow\frac{a^2+b^2}{c^2+d^2}=\frac{\left(a-b\right)^2}{\left(c-d\right)^2}\)

Đặt \(\frac{a}{b}=\frac{c}{d}=k\)

=>a=bk; c=dk

\(\frac{7a^2+3ab}{11a^2-8b^2}=\frac{7\cdot\left(bk\right)^2+3\cdot bk\cdot b}{11\cdot\left(bk\right)^2-8b^2}=\frac{7b^2k^2+3b^2\cdot k}{11b^2k^2-8b^2}=\frac{7k^2+3k}{11k^2-8}\)

\(\frac{7c^2+3cd}{11c^2-8d^2}=\frac{7\cdot\left(dk\right)^2+3\cdot dk\cdot d}{11\cdot\left(dk\right)^2-8d^2}=\frac{d^2\left(7k^2+3k\right)}{d^2\left(11k^2-8\right)}=\frac{7k^2+3k}{11k^2-8}\)

Do đó: \(\frac{7a^2+3ab}{11a^2-8b^2}=\frac{7c^2+3cd}{11c^2-8d^2}\) (ĐPCM)

Đặt \(\frac{a}{b}=\frac{c}{d}=k\)

=>a=bk; c=dk

\(\frac{7a^2+3ab}{11a^2-8b^2}=\frac{7\cdot\left(bk\right)^2+3\cdot bk\cdot b}{11\cdot\left(bk\right)^2-8b^2}=\frac{7b^2k^2+3b^2\cdot k}{11b^2k^2-8b^2}=\frac{7k^2+3k}{11k^2-8}\)

\(\frac{7c^2+3cd}{11c^2-8d^2}=\frac{7\cdot\left(dk\right)^2+3\cdot dk\cdot d}{11\cdot\left(dk\right)^2-8d^2}=\frac{d^2\left(7k^2+3k\right)}{d^2\left(11k^2-8\right)}=\frac{7k^2+3k}{11k^2-8}\)

Do đó: \(\frac{7a^2+3ab}{11a^2-8b^2}=\frac{7c^2+3cd}{11c^2-8d^2}\) (ĐPCM)

Ta có \(\frac{a}{c}=\frac{c}{b}\)=> \(\frac{a^2}{c^2}=\frac{c^2}{b^2}=\frac{a^2+b^2}{c^2+d^2}\)             (1)\

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

Từ 1,2 => đpcm

Ta có:

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

Ta lại có: 

\(\frac{a^2+c^2}{b^2+c^2}\Rightarrow\frac{a^2+ab}{b^2+ab}=\frac{a\left(a+b\right)}{b\left(a+b\right)}=\frac{a}{b}\)(đpcm)