Question

Cho tỉ lệ thức : \(\dfrac{a}{b}=\dfrac{c}{d}\). Chứng minh :

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

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

Answer 1

Đặt:

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

\(\dfrac{a+2b}{b}=\dfrac{bk+2b}{b}=\dfrac{b\left(k+2\right)}{b}=k+2\)

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

Vậy \(\dfrac{a+2b}{b}=\dfrac{c+2d}{d}\Rightarrowđpcm\)

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

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

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

Answer 2

a) ta có : \(\dfrac{a}{b}=\dfrac{c}{d}\Leftrightarrow ad=bc\Leftrightarrow ad+2bd=bc+2bd\)

\(\Leftrightarrow d\left(a+2b\right)=b\left(c+2d\right)\Leftrightarrow\dfrac{a+2b}{b}=\dfrac{c+2d}{d}\left(đpcm\right)\)

b) ta có : \(\dfrac{a}{b}=\dfrac{c}{d}\Leftrightarrow ad=bc\Leftrightarrow2ad=2bc\Leftrightarrow ad+ad=bc+bc\)

\(\Leftrightarrow ad-bc=bc-ad\Leftrightarrow ac+ad-bc-bd=ac+bc-ad-bd\)

\(\Leftrightarrow a\left(c+d\right)-b\left(c+d\right)=c\left(a+b\right)-d\left(a+b\right)\)

\(\Leftrightarrow\left(a-b\right)\left(c+d\right)=\left(c-d\right)\left(a+b\right)\Leftrightarrow\dfrac{a+b}{a-b}=\dfrac{c+d}{c-d}\left(đpcm\right)\)

Answer 3

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

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

\(\dfrac{a+2b}{b}=\dfrac{bk+2b}{b}=\dfrac{b\left(2+k\right)}{b}=k+2\left(1\right)\)

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

Từ (1) và (2)\(\Rightarrow\dfrac{a+2b}{b}=\dfrac{c+2d}{d}\)