A) \(\frac{a}{b}=\frac{c}{d}=t\Rightarrow a=bt,c=dt\)

\(\frac{a}{a+b}=\frac{bt}{bt+b}=\frac{t}{t+1},\frac{c}{c+d}=\frac{dt}{dt+d}=\frac{t}{t+1}\)

suy ra đpcm. 

\(\frac{a-b}{c-d}=\frac{bt-b}{dt-d}=\frac{b}{d},\frac{a+b}{c+d}=\frac{bt+b}{dt+d}=\frac{b}{d}\)

suy ra đpcm. 

B) \(\frac{a+3c}{b+3d}=\frac{a+c}{b+d}=\frac{\left(a+3c\right)-\left(a+c\right)}{\left(b+3d\right)-\left(b+d\right)}=\frac{2c}{2d}=\frac{c}{d}\)

\(\frac{a+3c}{b+3d}=\frac{a+c}{b+d}=\frac{\left(a+3c\right)-3\left(a+c\right)}{\left(b+3d\right)-3\left(b+d\right)}=\frac{-2a}{-2b}=\frac{a}{b}\)

suy ra đpcm. 

 a+b+c+d=0 
=>a+b=-(c+d) 
=> (a+b)^3=-(c+d)^3 
=> a^3+b^3+3ab(a+b)=-c^3-d^3-3cd(c+d) 
=> a^3+b^3+c^3+d^3=-3ab(a+b)-3cd(c+d) 
=> a^3+b^3+c^3+d^3=3ab(c+d)-3cd(c+d) ( vi a+b = - (c+d)) 
==> a^3 +b^^3+c^3+d^3==3(c+d)(ab-cd) (đpcm)

hey you, còn câu b,c?

ở đây có ai thích sơn tùng không ?

Ta có : \(ad=bc\)

=> \(\frac{a}{c}=\frac{b}{d}\)

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

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

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

mn ơi giúp mk với

a)a<b

=>a+c<b+c(1)

c<d

=>b+c<b+d(2)

Từ 1 và 2 =>a+c<b+d

b)a<b

=>ac<bc(1)

c<d

=>bc<bd(2)

Từ 1 và 2 =>ac<bd

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)