ai tra loi di chu

co biết làm đâu mà trả lời 

ta se tru moi ve di 1

2a+b+c+d/a -1 = a+b+c+d/b -1 =a+b+2c+d/c-1 = a+b+c+2d/d -1

a+b+c+d/a = a+b+c+d/b = a+b+c+d/c = a+b+c+d/d 

roi xet tung kha nang cua a+b+c+d=0

va a+b+c+d khac 0

ok nha 

byebye

love you

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

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

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

+) Xét \(a+b+c+d=0\)

Suy ra : 

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

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

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

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

Do đó : \(M=\frac{a+b}{c+d}+\frac{b+c}{d+a}+\frac{c+d}{a+b}+\frac{d+a}{c+b}\)

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

\(M=\left(-1\right)+\left(-1\right)+\left(-1\right)+\left(-1\right)\)

\(M=-4\)

+) Xét \(a+b+c+d\ne0\)

Áp dụng tính chất dãy tỉ số bằng nhau ta có : 

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

Do đó : 

\(\frac{a+b+c+d}{a}=4\)\(\Leftrightarrow\)\(a+b+c+d=4a\) \(\left(1\right)\)

\(\frac{a+b+c+d}{b}=4\)\(\Leftrightarrow\)\(a+b+c+d=4b\) \(\left(2\right)\)

\(\frac{a+b+c+d}{c}=4\)\(\Leftrightarrow\)\(a+b+c+d=4c\) \(\left(3\right)\)

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

Từ (1), (2), (3) và (4) suy ra \(4a=4b=4c=4d\) \(\left(=a+b+c+d\right)\)

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

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

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

Vậy \(M=-4\) hoặc \(M=4\)

Chúc bạn học tốt ~ 

Ta có : 

\(2a+2b+2c=by+cz+ax+cz+ax+by\)

\(\Leftrightarrow\)\(2\left(a+b+c\right)=2\left(ax+by+cz\right)\)

\(\Leftrightarrow\)\(a+b+c=ax+by+cz\)

+) \(a+b+c=ax+\left(by+cz\right)=ax+2a=a\left(x+2\right)\)

\(\Rightarrow\)\(\frac{1}{x+2}=\frac{a}{a+b+c}\) \(\left(1\right)\)

+) \(a+b+c=by+\left(ax+cz\right)=by+2b=b\left(y+2\right)\)

\(\Rightarrow\)\(\frac{1}{y+2}=\frac{b}{a+b+c}\) \(\left(2\right)\)

+) \(a+b+c=cz+\left(ax+by\right)=cz+2c=c\left(z+2\right)\)

\(\Rightarrow\)\(\frac{1}{z+2}=\frac{c}{a+b+c}\) \(\left(3\right)\)

Từ (1), (2) và (3) suy ra \(M=\frac{1}{x+2}+\frac{1}{y+2}+\frac{1}{z+2}\)

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

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

Vậy \(M=1\)

Chúc bạn học tốt ~ 

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

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

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

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

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

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

Vậy....

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