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' 😊😊😊

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đặt \(\frac{a}{b}=\frac{c}{d}=k\)

=> a=bk, c=dk  =>\(\frac{2a+5b}{3a-4b}=\frac{2bk+5b}{3bk-4b}=\frac{b\left(2k+5\right)}{b\left(3k-4\right)}=\frac{2k+5}{3k-4}\)(1)

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

từ (1)( 2)=> \(\frac{2a+5b}{3a-4b}=\frac{2c+5d}{3c-4d}\)

câu b c/m tg tự 

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

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

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

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

Xem ở Giúp tôi giải toán - Hỏi đáp, thảo luận về toán học - Học toán với OnlineMath

Đặt :

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

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

Theo đề ta có:

\(\dfrac{2a+5b}{3a-4b}=\dfrac{2c+5d}{3c-4d}\)

=> \(\dfrac{2a+5b}{3a-4b}-\dfrac{2c+5d}{3c-4d}\)

=> \(\dfrac{a+b}{a-b}-\dfrac{c+d}{c-d}\)(1)

Mà \(\dfrac{a}{b}=\dfrac{c}{d}\)

=> \(\dfrac{a}{c}=\dfrac{b}{d}\)(2)

=> \(\dfrac{a-b}{c-d}\) và \(\dfrac{a+b}{c+d}\)(3)

Từ (2) và (3) => \(\dfrac{a-b}{c-d}\) = \(\dfrac{a+b}{c+d}\) = \(\dfrac{a}{b}=\dfrac{c}{d}\)

=> \(\dfrac{a-b}{c-d}\) = \(\dfrac{a+b}{c+d}\)= > \(\dfrac{a-b}{a+b}\) = \(\dfrac{c-d}{c+d}\)

=> \(\dfrac{a+b}{a-b}=\dfrac{c+d}{c-d}\)= \(\dfrac{a+b}{a-b}-\dfrac{c+d}{c-d}\)(4)

Từ (1) và (4)

=> \(\dfrac{2a+5b}{3a-4b}=\dfrac{2c+5d}{3c-4d}\)( đpcm)

Đặt \(\dfrac{a}{b}=\dfrac{c}{d}=k\)

nên a=kb,c=kd.

Do đó: \(\dfrac{2a+5b}{3a-4b}=\dfrac{2kb+5b}{3kb-4b}=\dfrac{\left(2k+5\right)b}{\left(3k-4\right)b}=\dfrac{2k+5}{3k-4}\)

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

Suy ra \(\dfrac{2a+5b}{3a-4b}=\dfrac{2c+5d}{3c-4d}\)