Thiếu đề, bổ sung:
Cho: \(\frac{a}{b}=\frac{c}{d}\)C/m: \(\frac{2a+3b}{2a-3b}=\frac{2c+3d}{2c-3d}\)
Bài làm:
Đặt: \(\frac{a}{b}=\frac{c}{d}=k\)\(\left(k\ne0\right)\)
\(\Rightarrow\hept{\begin{cases}a=bk\\c=dk\end{cases}}\)
\(\frac{2a+3b}{2a-3b}=\frac{2bk+3b}{2bk-3b}=\frac{b\left(2k+3\right)}{b\left(2k-3\right)}=\frac{2k+3}{2k-3}\)\(\left(1\right)\)
\(\frac{2c+3d}{2c-3d}=\frac{2dk+3d}{2dk-3d}=\frac{d\left(2k+3\right)}{d\left(2k-3\right)}=\frac{2k+3}{2k-3}\)\(\left(2\right)\)
Từ \(\left(1\right)\)và \(\left(2\right)\)\(\Rightarrow\)\(\frac{2a+3b}{2a-3b}=\frac{2c+3d}{2c-3d}\)\(\left(đpcm\right)\)
Cách khác nhanh hơn bạn Lạc :
Ta có : \(\frac{a}{b}=\frac{c}{d}\)
<=> \(\frac{a}{c}=\frac{b}{d}\)( t/c tỉ lệ thức)
<=> \(\frac{2a}{2c}=\frac{3b}{3d}\)
<=> \(\frac{2a}{2c}=\frac{3b}{3d}=\frac{2a+3b}{2c+3d}=\frac{2a-3b}{2c-3d}\)(t/c DTSBN)
<=> \(\frac{2a+3b}{2a-3b}=\frac{2c+3d}{2c-3d}\)(t/c tlt)