a/ Đặt :

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

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

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

Từ \(\left(1\right)+\left(2\right)\Leftrightarrowđpcm\)

b/ Đặt :

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

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

\(VP=\dfrac{c+d}{d}=\dfrac{dk+d}{d}=\dfrac{d\left(k+1\right)}{d}=\dfrac{k+1}{d}\left(2\right)\)

Từ \(\left(1\right)+\left(2\right)\Leftrightarrowđpcm\)

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

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

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

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

\(2d\times2a=2b\times2c\)

\(ad=bc\)

\(\frac{a}{c}=\frac{b}{d}\left(\text{đ}pcm\right)\)

\(\Leftrightarrow\frac{2a+b+c+d}{a}-1=\frac{a+2b+c+d}{b}-1=\frac{a+b+2c+d}{c}-1=\frac{a+b+c+2d}{d}-1\)

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

\(\text{Th}1:a+b+c+d=0\Rightarrow\hept{\begin{cases}a+b=-\left(c+d\right)\\b+c=-\left(a+d\right)\end{cases}}\)

\(M=\frac{-\left(c+d\right)}{c+d}+\frac{-\left(a+d\right)}{a+d}+\frac{c+d}{-\left(c+d\right)}+\frac{d+a}{-\left(a+d\right)}=-4\)

\(\text{th}2:a+b+c+d\ne0\Rightarrow a=b=c=d\)

\(\Leftrightarrow M=1+1+1+1=4\)

Vậy....

p/s: đầu tiên nhớ ghi lại cái đề nha :)) 

Trừ cả 4 vế cho 1 ta có:\(\frac{a+b+c+d}{a}=\frac{a+b+c+d}{b}=\frac{a+b+c+d}{c}=\frac{a+b+c+d}{d}\)

Áp dụng t/c của dãy tỉ số bằng nhau ta có:

\(\frac{a+b+c+d}{a}=\frac{a+b+c+d}{b}=\frac{a+b+c+d}{c}=\frac{a+b+c+d}{d}=\frac{4\left(a+b+c+d\right)}{a+b+c+d}=4\)

Suy ra :

            a+b+c+d=4a=4b=4c=4d hay a=b=c=d

Do đó:

M=\(\frac{a+b}{c+d}+\frac{b+c}{d+a}+\frac{c+d}{a+b}+\frac{a+d}{b+c}=4\)

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

\(\Leftrightarrow1+\frac{a+b+c+d}{a}=\frac{a+b+c+d}{b}=\frac{a+b+c+d}{c}=\frac{a+b+c+d}{d}\left(ĐK:a,b,c,d\ne0\right)\)

\(TH1:a+b+c+d=0\)

\(\Rightarrow\hept{\begin{cases}a+b=-\left(c+d\right)\\b+c=-\left(a+d\right)\end{cases}}\)

\(\Rightarrow M=\frac{-\left(c+d\right)}{c+d}+\frac{-\left(b+c\right)}{b+c}+\frac{c+d}{-\left(c+d\right)}+\frac{d+a}{-\left(a+d\right)}=-1.4=-4\)

\(TH2:a+b+c+d=0\)

\(\Rightarrow a=b=c=d\)

\(\Rightarrow M=\frac{a+a}{a+a}+\frac{b+b}{b+b}+\frac{c+c}{c+c}+\frac{d+d}{d+d}=1.4=4\)

Vậy M=-4 hay M=4

p/s: =.= bn sử dụng công thức cho dễ đọc tí nhé :>

sorry >:

dòng 2:

\(1+\frac{a+b+c+d}{a}=1+\frac{a+b+c+d}{b}=1\)\(=1+\frac{a+b+c+d}{c}=1+\frac{a+b+c+d}{d}\)

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

I donnt no

\(\frac{2a+b+c+d}{a}-1=\frac{a+2b+c+d}{b}-1=\frac{a+b+2c+d}{c}-1=\frac{a+b+c+2d}{d}-1\)

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

\(\Rightarrow a=b=c=d\)

\(M=1+1+1+1=4\)

a+b+c+d=22