Question

\(Cho\frac{a}{b}=\frac{c}{d}\left(\ne1\right).\)

Chứng minh rằng:  \(\frac{3a^2+5ab}{a^2-b^2}\)=\(\frac{3c^2+5cd}{c^2-d^2}\)

 

Answer 1

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

=> \(\frac{a}{c}=\frac{b}{d}\)

=> \(\frac{a^2}{c^2}=\frac{b^2}{d^2}=\frac{3a^2}{3c^2}=\frac{ab}{cd}=\frac{5ab}{5cd}=\frac{a^2-b^2}{c^2-d^2}\)

Áp dụng tính chất dãy tỉ số bằng nhau ta có:

\(\frac{3a^2}{3c^2}=\frac{5ab}{5cd}=\frac{a^2-b^2}{c^2-d^2}=\frac{3a^2+5ab}{3c^2+5cd}\)

=> \(\frac{a^2-b^2}{c^2-d^2}=\frac{3a^2+5ab}{3c^2+5cd}\)

=> \(\frac{3a^2+5ab}{a^2-b^2}=\frac{3c^2+5cd}{c^2-d^2}\)

=> Đpcm

Answer 2

đặt \(\frac{a}{b}=\frac{c}{d}=k\)

=>a=bk

    c=dk

bạn thay vào rùi làm tiếp

Answer 3

Đặt \(\frac{a}{b}=\frac{c}{d}=k\)