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

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

Tương tự

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

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

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

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

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

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

\(\Rightarrow\frac{a+b}{c+d}+\frac{b+c}{d+a}+\frac{c+d}{a+b}+\frac{d+a}{b+c}=1+1+1+1=4\)

<br class="Apple-interchange-newline"><div id="inner-editor"></div>2a+b+c+da =a+2b+c+db =a+b+2c+dc =a+b+c+2dd =2a+b+c+d+a+2b+c+d+a+b+2c+d+a+b+c+2ca+b+c+d =4

=>2a+b+c+d=4a

=>2a=b+c+d

Tương tự ta có:2b=a+c+d

2c=a+b+d

2d=a+b+c

=>2a+2b=b+c+d+a+c+d=>a+b+2c+2d

=>a+b=2c+2d

=>a+b/c+d=2

Tương tự ta có:b+c/d+a=2

c+d/a+b=2

d+a/b+c=2

=>M=2+2+2+2=8

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Câu hỏi của vuong hien duc

Ta có: \(\frac{2012a+b+c+d}{a}-2011=\frac{a+2012b+c+d}{b}-2011=\frac{a+b+2012c+d}{c}-2011\)

\(=\frac{a+b+c+2012d}{d}-2011\)

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

+) Xét \(a+b+c+d=0\)

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

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

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

\(=\left(-1\right)+\left(-1\right)+\left(-1\right)+\left(-1\right)=-4\)

+) Xét \(a+b+c+d\) khác 0 \(\Rightarrow a=b=c=d\)

\(M=\frac{a+b}{c+d}+\frac{b+c}{d+a}+\frac{c+d}{a+b}+\frac{d+a}{b+c}=1+1+1+1=4\)

Vậy...

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

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

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

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

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

Xét \(a+b+c+d=0\) ta có : 

\(a+b=-c-d;b+c=-a-d;c+d=-a-b;d+a=-b-c\)

\(\Rightarrow A=\frac{a+b}{-a-b}+\frac{b+c}{-b-c}+\frac{c+d}{-c-d}+\frac{d+a}{-b-c}=-1-1-1-1=-4\)

Xét \(a+b+c+d\ne0\) ta có : \(a=b=c=d\)

\(\Rightarrow M=\frac{a+a}{a+a}+\frac{a+a}{a+a}+\frac{a+a}{a+a}+\frac{a+a}{a+a}=1+1+1+1=4\)

Xét \(a+b+c+d=0\) thì ta có dãy tỷ số là đúng.

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

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

Xét \(a+b+c+d\ne0\)thì ta có:

\(\frac{2015a+b+c+d}{a}=\frac{a+2015b+c+d}{b}=\frac{a+b+2015c+d}{c}=\frac{a+b+c+2015d}{d}=\frac{2018\left(a+b+c+d\right)}{a+b+c+d}=2018\)

Lấy 2 cái đầu cộng với nhau ta được:

\(\frac{2016\left(a+b\right)+2\left(c+d\right)}{a+b}=2018\)

\(\Leftrightarrow\frac{c+d}{a+b}=\frac{2018-2016}{2}=1\)

Tương tự ta cũng có:

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

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

Ta có:

\(\frac{2012a+b+c+d}{a}-2011=\frac{a+2012b+c+d}{b}-2011=\frac{a+b+2012c+d}{c}-2011\)\(=\frac{a+b+c+2012d}{d}\)

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

+) Xét \(a+b+c+d=0\)

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

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

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

\(=\left(-1\right)+\left(-1\right)+\left(-1\right)+\left(-1\right)=-4\)

+) Xét \(a+b+c+d\ne0\Rightarrow a=b=c=d\)

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

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

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

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

Vậy ...

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Câu hỏi của Đào Ngọc Bảo Anh - Toán lớp 7 - Học toán với OnlineMath

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

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

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

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

Xét \(a+b+c+d=0\)

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

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

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

Xét \(a+b+c+d\ne0\Rightarrow a=b=c=d\)

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

Vậy M=-4 hoặc M=4