Đặt : \(\frac{a}{b}=\frac{c}{d}=k\Rightarrow\begin{cases}a=bk\\c=dk\end{cases}\)

\(\frac{2a+3b}{3a-4b}=\frac{2bk+3b}{3bk-4b}=\frac{b\left(2k+3\right)}{b\left(3k-4\right)}=\frac{2k+3}{3k-4}\)

\(\frac{2c+3d}{3c-4d}=\frac{2dk+3d}{3dk-4d}=\frac{d\left(2k+3\right)}{d\left(3k+4\right)}=\frac{2k+3}{3k-4}\)

Vậy \(\frac{2a+3b}{3a-4b}=\frac{2c+3d}{3c-4d}\) \(\left(đpcm\right)\)

Giải:

Đặt \(\frac{a}{b}=\frac{c}{d}=k\)

Ta có:

a = b.k

c = d.k

Theo bài ra ta có:
\(\frac{2a+3b}{3a-4b}=\frac{2.b.k+3.b}{3.b.k-4.b}=\frac{b\left(2.k+3\right)}{b.\left(3.k-4\right)}=\frac{2.k+3}{3.k-4}\)   (1)

\(\frac{2c+3d}{3c-4d}=\frac{2.d.k+3d}{3.d.k-4d}=\frac{d.\left(2.k+3\right)}{d.\left(3.k-4\right)}=\frac{2.k+3}{3.k-4}\)   (2)

Từ (1) và (2) suy ra \(\frac{2a+3b}{3a-4d}=\frac{2c+3d}{3c-4d}\Rightarrowđpcm\)

Đặt \(\frac{a}{b}=\frac{c}{d}=k\Rightarrow\hept{\begin{cases}a=bk\\c=dk\end{cases}}\)

\(\frac{2a+3b}{3a-4b}=\frac{2bk+3b}{3bk-4b}=\frac{b\left(2k+3\right)}{b\left(3k-4\right)}=\frac{2k+3}{3k-4}\)

\(\frac{2c+3d}{3c-4d}=\frac{2dk+3d}{3dk-4d}=\frac{d\left(2k+3\right)}{d\left(3k-4\right)}=\frac{2k+3}{3k-4}\)

Vậy \(\frac{2a+3b}{3a-4b}=\frac{2c+3d}{3c-4d}\)(đpcm)

k có giả thuyết thì sao chứng minh 

a)  đặt \(\frac{a}{b}=\frac{c}{d}=k\Leftrightarrow a=b.k;c=d.k\)

          \(\frac{3a+2c}{3b+2d}=\frac{3b.k+2.d.k}{3b+2d}=\frac{k\left(3b+2d\right)}{3b+2d}=k\)

    b)          bó tay