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Giải:

Ta có: \(\frac{a}{b}=\frac{c}{d}\Rightarrow\frac{a}{c}=\frac{b}{d}\)

Áp dụng tính chất dãy tỉ số bằng nhau ta có:
\(\frac{a}{c}=\frac{b}{d}=\frac{2a}{2c}=\frac{5b}{5d}=\frac{2a+5b}{2c+5d}\)

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

\(\Rightarrow\frac{2a+5b}{2c+5d}=\frac{3a-4b}{3c-4d}\left(=\frac{a}{c}\right)\)

\(\Rightarrow\frac{2a+5b}{3a-4b}=\frac{2c+5d}{3c-4d}\left(đpcm\right)\)

Vậy...

Gọi a/b=c/d=k nên a=bk;c=dk

=>2a+5b/3a-4b=2bk+5b/3bk-4b=b(2k+5)/b(3k-4)=2k+5/3k-4(1)

=>2c+5d/3c-4d=2dk+5d/3dk-4d=d(2k+5)/d(3k-4)=2k+5/3k-4(2)

Từ (1);(2) =>2a+5b/3a-4b=2c+5d/3c-4d

Thank Đỗ Lê Tú Linh n' 😊😊😊

😀😀😀

a/ Đặt :

\(\dfrac{a}{b}=\dfrac{c}{d}=k\)

\(\Leftrightarrow\left\{{}\begin{matrix}a=bk\\c=dk\end{matrix}\right.\)

Ta có :

\(VT=\dfrac{a-b}{a+b}=\dfrac{bk-b}{bk+b}=\dfrac{b\left(k-1\right)}{b\left(k+1\right)}=\dfrac{k-1}{k+1}\left(1\right)\)

\(VP=\dfrac{c-d}{c+d}=\dfrac{dk-d}{dk+d}=\dfrac{d\left(k-1\right)}{d\left(k+1\right)}=\dfrac{k-1}{k+1}\left(2\right)\)

Từ \(\left(1\right)+\left(2\right)\Leftrightarrowđpcm\)

b/ Đặt :

\(\dfrac{a}{b}=\dfrac{c}{d}=k\) \(\Leftrightarrow\left\{{}\begin{matrix}a=bk\\c=dk\end{matrix}\right.\)

Ta có :

\(VT=\dfrac{2a-5b}{3a+4b}=\dfrac{2bk-5b}{3bk+4b}=\dfrac{b\left(2k-5\right)}{b\left(3k+4\right)}=\dfrac{2k-5}{3k+4}\left(1\right)\)

\(VP=\dfrac{2c-5d}{3c+4d}=\dfrac{2dk-5d}{3dk+4d}=\dfrac{d\left(2k-5\right)}{d\left(3k+4\right)}=\dfrac{2k-5}{3k+4}\left(2\right)\)

Từ \(\left(1\right)+\left(2\right)\Leftrightarrowđpcm\)

Ta đặt:\(\frac{a}{b}=\frac{c}{d}=k\Rightarrow a=bk;c=dk\)

Khi đó: \(\frac{2a+5b}{3a-4b}=\frac{2bk+5b}{3bk-4b}=\frac{b\left(2k+5\right)}{b\left(3k-4\right)}=\frac{2k+5}{3k-4}\)

            \(\frac{2c+5d}{3c-4d}=\frac{2dk+5d}{3dk-4d}=\frac{d\left(2k+5\right)}{d\left(3k-4\right)}=\frac{2k+5}{3k-4}\)

\(\Rightarrow\frac{2a+5b}{3a-4b}=\frac{2c+5d}{3c-4d}\left(=\frac{2k+5}{3k-4}\right)\)