\(\frac{a}{b}=\frac{c}{d}\)

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

\(\Leftrightarrow8ac-10ad+12bc-15bd=8ac+12ad-10bc-15bd\)

\(\Leftrightarrow-10ad+12bc=12ad-10bc\)

\(\Leftrightarrow\left(-10ad+12bc\right)+\left(-12bc-12ad\right)=\left(12ad-10bc\right)+\left(-12bc-12ad\right)\)

\(\Leftrightarrow22bc=22ad\)

vì a/b=c/d nên => a/c=b/d

đặt a/c=b/d =k thì => a=ck ; b= dk 

thay a=ck và b=dk vào 2a-3b/4a+5b có 

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

từ đay suy ra 2a-3b/4a+5b=2c-3d/4c+5d 

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...

Ta có: \(\frac{a}{b}=\frac{c}{d}\)

\(\frac{2a-3b}{5b}=\frac{2c-3d}{5d}\\ \Leftrightarrow\frac{d}{b}=\frac{2c-3d}{2a-3b}\)

Áp dụng TCDTSBN ta có:

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

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

Áp dụng TCDTSBN ta có:

\(\frac{d}{b}=\frac{c-d}{a-b}=\frac{d+c-d}{b+a-b}=\frac{c}{a}\\ \Rightarrow\frac{d}{b}=\frac{c}{a}\\ \Leftrightarrow\frac{a}{b}=\frac{c}{d}\left(Theo.gt\right)\\ \Rightarrow dpcm\)