mn ơi giúp mk với

Ta có : \(ad=bc\)

=> \(\frac{a}{c}=\frac{b}{d}\)

\(ADTCDTSBN,tađược\):
\(\frac{a}{c}=\frac{b}{d}=\frac{a-b}{c-d}=\frac{a+b}{c+d}\)

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

=> \(\frac{a-b}{a+b}=\frac{c-d}{c+d}\left(đpcm\right)\)

CÁC CẬU ƠI GIÚP MIK VS!!!!!!

Đặt \(\frac{a}{b}=\frac{c}{d}=k\left(k\ne0\right)\Rightarrow a=kb;c=kd\)

\(\frac{a+b}{b}=\frac{kb+b}{b}=\frac{b\left(k+1\right)}{b}=k+1\)

\(\frac{c+d}{d}=\frac{kd+d}{d}=\frac{d\left(k+1\right)}{d}=k+1\)

Vậy: \(\frac{a+b}{b}=\frac{c+d}{d}\left(=k+1\right)\)

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

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

Ta có: \(\dfrac{a-b}{b}=\dfrac{bk-b}{b}=k-1\)

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

Do đó: \(\dfrac{a-b}{b}=\dfrac{c-d}{d}\)

Đặt \(\dfrac{a}{b}=\dfrac{c}{d}=k\)

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

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

\(\Leftrightarrow\dfrac{bk+dk}{b+d}=\dfrac{bk-dk}{b-d}\)

hay k=k(đúng)

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

\(=-\left(b+c\right)^2+\left(a+d\right)^2\)   ( 1 )

\(\left(a+b-c-d\right)\left(a-b+c-d\right)=\left(b-c\right)^2-\left(a-d\right)^2\)    ( 2 )

Từ ( 1 ) và ( 2 ) suy ra 

\(b^2+2bc+c^2-a^2-2ad-d^2=a^2-2ad+d^2-b^2+2bc-c^2\)

\(4ad=4ac\Rightarrow ad=bc\)

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