Question

cho \(\dfrac{a}{b}\)=\(\dfrac{c}{d}\) C/M \(\dfrac{7a^2+5ac}{7a^2-5ac}\)= \(\dfrac{7b^2+5bd}{7b^2-5bd}\)

Answer 1

Với \(\dfrac{a}{b}=\dfrac{c}{d}\)

=> \(\dfrac{a}{b}.\)\(\dfrac{c}{d}=\dfrac{ac}{bd}=\dfrac{aa}{bb}=\dfrac{a^2}{b^2}\)
Ta có : \(\dfrac{a^2}{b^2}=\dfrac{ac}{bd}\)

=> \(\dfrac{7a^2}{7b^2}=\dfrac{5ac}{5bd}\)

Áp dụng t/c dãy tỉ số bằng nhau:

\(\dfrac{7a^2}{7b^2}=\dfrac{5ac}{5bd}=\dfrac{7a^2+5ac}{7b^2+5bd}=\dfrac{7a^2-5ac}{7b^2-5bd}\) (1)

Từ (1) => \(\dfrac{7a^2+5ac}{7a^2-5ac}=\dfrac{7b^2-5bd}{7b^2-5bd}\) (ĐPCM)