Chứng minh rằng nếu a/ b = c /d thì b) (7a^2 + 3ab)/( 11a^2 − 8b^2) = (7c^2 + 3 cd)/( 11c^2 − 8d^2) .

b) Ta có:

• \(\frac{{7{a^2} + 3ab}}{{11{a^2} - 8{b^2}}} = \frac{{7{k^2}{b^2} + 3k{b^2}}}{{11{k^2}{b^2} - 8{b^2}}} = \frac{{{b^2}\left( {7{k^2} + 3k} \right)}}{{{b^2}\left( {11{k^2} - 8} \right)}} = \frac{{7{k^2} + 3k}}{{11{k^2} - 8}}{\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} \left( 3 \right)\)

• \(\frac{{7{c^2} + 3cd}}{{11{c^2} - 8{d^2}}} = \frac{{7{k^2}{d^2} + 3k{d^2}}}{{11{k^2}{d^2} - 8{d^2}}} = \frac{{{d^2}\left( {7{k^2} + 3k} \right)}}{{{d^2}\left( {11{k^2} - 8} \right)}} = \frac{{7{k^2} + 3k}}{{11{k^2} - 8}}{\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} \left( 4 \right)\)

Từ (3) và (4) suy ra: \(\frac{{7{a^2} + 3ab}}{{11{a^2} - 8{b^2}}} = \frac{{7{c^2} + 3cd}}{{11{c^2} - 8{d^2}}}\).