Cho \(\frac{a}{b} = \frac{c}{d}\). Chứng minh rằng \(\frac{{7{a^2} + 3ab}}{{11{a^2} - 8{b^2}}} = \frac{{7{c^2} + 3cd}}{{11{c^2} - 8{d^2}}}\).

Từ \(\frac{a}{b} = \frac{c}{d} \Rightarrow \frac{a}{c} = \frac{b}{d}\)

Suy ra \(\frac{{{a^2}}}{{{c^2}}} = \frac{{{b^2}}}{{{d^2}}} = \frac{{ab}}{{cd}} = \frac{{7{a^2}}}{{7{c^2}}} = \frac{{8{b^2}}}{{8{d^2}}} = \frac{{3ab}}{{3cd}} = \frac{{11{a^2}}}{{11{c^2}}}\).

Áp dụng tính chất dãy tỉ số bằng nhau cho \(\frac{{{a^2}}}{{{c^2}}} = \frac{{11{a^2}}}{{11{c^2}}} = \frac{{8{b^2}}}{{8{d^2}}}\), ta được:

\(\frac{{{a^2}}}{{{c^2}}} = \frac{{11{a^2}}}{{11{c^2}}} = \frac{{8{b^2}}}{{8{d^2}}} = \frac{{11{a^2} - 8{b^2}}}{{11{c^2} - 8{d^2}}}\)          (1)

Áp dụng tính chất dãy tỉ số bằng nhau cho \(\frac{{{a^2}}}{{{c^2}}} = \frac{{7{a^2}}}{{7{c^2}}} = \frac{{3ab}}{{3cd}}\), ta được:

\(\frac{{{a^2}}}{{{c^2}}} = \frac{{7{a^2}}}{{7{c^2}}} = \frac{{3ab}}{{3cd}} = \frac{{7{a^2} + 3ab}}{{7{c^2} + 3cd}}\)  (2)

Từ (1) và (2) suy ra: \(\frac{{11{a^2} - 8{b^2}}}{{11{c^2} - 8{d^2}}} = \frac{{7{a^2} + 3ab}}{{7{c^2} + 3cd}}\).

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