Olm chào em, em cần chứng minh điều gì với dãy tỉ số này. em nhỉ?

Bằng nhau rồi thì bạn định làm gì với nó?

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

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

Tương tự

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

Nếu a + b + c + d = 0 => a + b = -(c + d) ; (b + c) = -(a + d) ; c + d = -(a+b) ; d + a = -(b + c)

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

Vậy M = 4 hoặc M = -4

Ta có:\(\frac{2a+b+c+d}{a}=\frac{a+2b+c+d}{b}=\frac{a+b+2c+d}{c}=\frac{a+b+c+2d}{d}=\frac{2a+b+c+d+a+2b+c+d+a+b+2c+d+a+b+c+2c}{a+b+c+d}=4\)

=>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

Moon Light sai rồi bn nhé

Cộng vào bằng 5 nhé

Đề bài: Cho dãy tỉ số bằng nhau:

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

Tính giá trị biểu thức \(M=\frac{a+b}{c+d}+\frac{b+c}{d+a}+\frac{c+d}{a+b}+\frac{d+a}{b+c}\)

Bài làm

Cùng trừ mỗi tỉ số trên đi 1 đơn vị ta được:

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

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

Từ đây ta suy ra 2 trường hợp:

+ Trường hợp 1:

Nếu a + b + c + d \(\notin0\) => a = b = c = d

=> M = 1 + 1 + 1 + 1 = 1 . 4 = 4

+ Trường hợp 2:

Nếu a + b + c + d = 0 thì

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

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

Do đó: M = ( -1 ) + ( - 1 ) + ( - 1 ) + ( - 1) = -4

Bạn ơi giải thích cho mình chỗ a+b= -(c+d) được k? Mình vẫn không hiểu lắm!

2a+b+c+da=a+2b+c+db=a+b+2c+dc=a+b+c+2dd

↔a+a+b+c+da=a+b+b+c+db=a+b+c+c+dc=a+b+c+d+dd

↔a+b+c+da+1=a+b+c+db+1=a+b+c+dc+1=a+b+c+dd+1

↔a+b+c+da=a+b+c+db=a+b+c+dc=a+b+c+dd

đến đây em xét 2 TH:

a+b+c+d≠0

a+b+c+d=0

__________________

Ta có:\(\frac{2a+b+c+d}{a}=\frac{a+2b+c+d}{b}=\frac{a+b+2c+d}{c}=\frac{a+b+c+2d}{d}=\frac{2a+b+c+d+a+2b+c+d+a+b+2c+d+a+b+c+2c}{a+b+c+d}=4\)

=>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

ban co the giai tuong tan ra ko minh ko hiu