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từ tỉ lệ thức đã cho
=>(3a+4b)(5c-6d)=(3c+4d)(5a-6b)
=>15ac-18ad+20bc-24bd=15ac+20ad-18bc-24bd
=>-18ad+20bc=20ad-18bc
=>-18ad-20ad=-18bc-20bc
=>-38ad=-38bc
=>ad=bc
=>a/b=c/d
=>
\(\dfrac{3a+4b}{5a-6b}=\dfrac{3c+4d}{5c-6d}\)
=> \(\dfrac{3a+4b}{3c+4d}=\dfrac{5a-6b}{5c-6d}\)
ta có
\(\dfrac{3a+4b}{3c+4d}=\dfrac{3a}{3c}=\dfrac{4b}{4d}=\dfrac{a}{c}=\dfrac{b}{d}=>\dfrac{a}{b}=\dfrac{c}{d}\)(đpcm)
Ta có:
\(\dfrac{3a+4b}{5a-6b}=\dfrac{3c+4d}{5c-6d}\)
\(\Leftrightarrow\left(3a+4b\right)\left(5c-6d\right)=\left(3c+4d\right)\left(5a-6b\right)\)
\(\Rightarrow15ac-18ad+20bc-24bd=15ac-18bc+20ad-24bd\)
\(\Rightarrow15ac-15ac-18ad-20ad=-24bd+24bd-18bc-20bc\)
\(\Rightarrow-38ad=-38bc\)
\(\Rightarrow ad=bc\)
\(\Rightarrow\dfrac{a}{b}=\dfrac{c}{d}\)
Theo đề ra, ta có:
\(ad=bc\Rightarrow\frac{a}{b}=\frac{c}{d}=N\Rightarrow a=bN;c=dN\)
Xét \(\frac{3a+4b}{5a+6b}=\frac{3bN+4b}{5bN+6b}=\frac{b\left(3N+4\right)}{b\left(5N+6\right)}=\frac{3N+4}{5N+6}\)
\(\frac{3c+4d}{5c+6d}=\frac{3dN+4d}{5dN+6d}=\frac{d\left(3N+4\right)}{d\left(5N+6\right)}=\frac{3N+4}{5N+6}\)
Do vậy: \(\frac{3a+4b}{5a+6b}=\frac{3c+4d}{5c+6d}\left(đpcm\right)\)