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

vì a+b+c+d=0 nên a+b=0-c-d=-(c+d)

Bài làm đây nha bạn

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

hình như cộng lại bằng 5 chứ bạn

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

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

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

=>a=b=c=d

=>a+b=c+d; b+c=d+a

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

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

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

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

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

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

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

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

+) Nếu \(a+b+c+d=0\Rightarrow a+b=-\left(c-d\right);b+c=-\left(d+a\right)\)

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

Ta có:

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

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

    \(=\dfrac{a+b+c+2d}{d}-1\)

⇔ \(\dfrac{a+b+c+d}{a}=\dfrac{a+b+c+d}{b}=\dfrac{a+b+c+d}{c}=\dfrac{a+b+c+d}{d}\)

Nếu a+b+c+d=0

⇒a+b=−(c+d);c+b=−(a+d);c+d=−(a+b);a+d=−(c+b)

Thay vào M, ta có:

\(M=\dfrac{a+b}{-\left(a+b\right)}=\dfrac{b+c}{-\left(b+c\right)}=\dfrac{c+d}{-\left(c+d\right)}=\dfrac{a+d}{-\left(a+d\right)}=-1\)

Nếu a+b+c+d ≠0

⇒ \(a=b=c=d\)

Thay vào M, ta có

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

\(\text{Cùng trừ mỗi tỉ số trên 1 đơn vị ta được:}\)

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

\(\text{Từ đây ta suy ra 2 trường hợp:}\)

\(\text{Trường hợp 1:}\)

\(\text{Nếu }a+b+c+d\notin0\Rightarrow a=b=c=d\)

\(\Rightarrow M=1+1+1+1=1.4=4\)

\(\text{Trường hợp 2:}\)

\(\text{Nếu }a+b+c+d=0\text{ thì:}\)

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

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

\(\text{Do đó }M=\left(-1\right)+\left(-1\right)+\left(-1\right)+\left(-1\right)=-4\)