Question

cho \(\frac{a}{b}=\frac{c}{d}\)b,d khác 0 CMR \(\frac{2a+3b}{a+b}=\frac{2c+3d}{c+d}\)

Answer 1

nhanh nhanh nha nha nha mai nộp rùi nha nha

Answer 2

Đặt \(\frac{a}{b}=\frac{c}{d}=k\Rightarrow a=bk,c=dk.\)

Ta có: \(\frac{2a+3b}{a+b}=\frac{2bk+3b}{bk+b}=\frac{b\left(2k+3\right)}{b\left(k+1\right)}=\frac{2k+3}{k+1}\left(1\right)\)

và \(\frac{2c+3d}{c+d}=\frac{2dk+3d}{dk+d}=\frac{d\left(2k+3\right)}{d\left(k+1\right)}=\frac{2k+3}{k+1}\left(2\right)\)

Từ (1) và (2) ta suy ra: 2a+3ba+b=2c+3dc+d (đpcm)