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

Xét \(VT=\frac{a+2b}{a-2b}=\frac{bk+2b}{bk-2b}=\frac{b\left(k+2\right)}{b\left(k-2\right)}=\frac{k+2}{k-2}\left(1\right)\)

Xét \(VP=\frac{c+2d}{c-2d}=\frac{dk+2d}{dk-2d}=\frac{d\left(k+2\right)}{d\left(k-2\right)}=\frac{k+2}{k-2}\left(2\right)\)

Từ (1) và (2) =>Đpcm

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:

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

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

sao khó thế

cho \(\frac{a}{b}=\frac{c}{d}\)chứng minh rằng giả thiết trên nha(các gt đều có nghĩa)

ta có \(\frac{a}{b}\)=\(\frac{c}{d}\)đặt k ta có\(\hept{\begin{cases}a=b.k\\c=d.k\end{cases}}\)

vậy ta có\(\hept{\begin{cases}\frac{2\left(b.k\right)+2b}{2\left(b.k\right)-2b}=\frac{2b.k+2b}{2b.k-2b}=\frac{2b.\left(k+1\right)}{2b.\left(k-1\right)}=\frac{k+1}{k-1}\left(1\right)\\\frac{2\left(d.k\right)+2d}{2\left(d.k\right)-2d}=\frac{2d.k+2d}{2d.k-2d}=\frac{2d.\left(k+1\right)}{2d.\left(k-1\right)}=\frac{k+1}{k-1}\left(2\right)\end{cases}}\)

từ (1) và (2) ta được

=>\(\frac{k+1}{k-1}=\frac{k+1}{k-1}\)       vậy\(\frac{2a+2b}{2a-2b}\)=\(\frac{2c+2d}{2c-2d}\)(điều phải chứng minh)

Đặt:

\(\dfrac{a}{b}=\dfrac{c}{d}=k\)

\(\Rightarrow\left\{{}\begin{matrix}a=bk\\c=dk\end{matrix}\right.\)

\(\Rightarrow\dfrac{a+2b}{c+2d}=\dfrac{bk+2b}{dk+2d}=\dfrac{b\left(k+2\right)}{d\left(k+2\right)}=\dfrac{b}{d}\)

\(\Rightarrow\dfrac{a-2b}{c-2d}=\dfrac{bk-2b}{dk-2d}=\dfrac{b\left(k-2\right)}{d\left(k-2\right)}=\dfrac{b}{d}\)

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

ta có : \(\dfrac{a}{b}=\dfrac{c}{d}\Leftrightarrow ad=bc\Leftrightarrow4ad=4bc\Leftrightarrow2ad+2ad=2bc+2bc\)

\(\Leftrightarrow2ad-2bc=2bc-2ad\Leftrightarrow ac+2ad-2bc-4bd=ac+2bc-2ad-4bd\)

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