Question

[Nguyễn Trần Như Hằng]

c/m;

Từ a/b=c/d : 2a+3b/3a-4b = 2c+3d/3c-4d

Answer 1

[Võ Đông Anh Tuấn]

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

Answer 2

[Nguyễn Huy Tú]

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