\(\dfrac{a}{b}=\dfrac{c}{d}\Rightarrow ad=bc\)
1.
giả sử điều đó đúng thì:
\(c\left(b+a\right)=a\left(c+d\right)\\ bc+ca=ac+ad\Rightarrow bc+ca=ca+bc\left(đúng\right)\)
\(\Rightarrow\dfrac{a}{b+a}=\dfrac{c}{d+c}\)
2.
\(\dfrac{a-2b}{b}=\dfrac{c-2d}{d}\\ \dfrac{a-b}{b}-1=\dfrac{c-d}{d}-1\\ \dfrac{a-b}{b}=\dfrac{c-d}{d}\\ \left(a-b\right)d=\left(c-d\right)b\\ ad-bd=bc-bd\\ \Rightarrow ad-bd=ad-bd\left(đúng\right)\)
\(\Rightarrow\dfrac{a-2b}{b}=\dfrac{c-2d}{d}\) cũng đúng
1)
\(\dfrac{a}{b}=\dfrac{c}{d}\Leftrightarrow ad=bc\)
\(\dfrac{a}{b+a}=\dfrac{c}{c+d}\Leftrightarrow a\left(c+d\right)=c\left(b+a\right)\)
\(\Leftrightarrow ac+ad=bc+ac\Leftrightarrow ad=bc\)
\(\Leftrightarrow\dfrac{a}{b+a}=\dfrac{c}{c+d}\)
2)
\(\dfrac{a}{b}=\dfrac{c}{d}\)
\(\Leftrightarrow\dfrac{a}{b}-2=\dfrac{c}{d}-2\)
\(\Leftrightarrow\dfrac{a}{b}-\dfrac{2b}{b}=\dfrac{c}{d}-\dfrac{2d}{d}\)
\(\Leftrightarrow\dfrac{a-2b}{b}=\dfrac{c-2d}{d}\rightarrowđpcm\)
