Question

[okokok]

Cho dãy tỉ số bằng nhau \(\dfrac{a}{b+c+d}=\dfrac{b}{a+c+d}=\dfrac{c}{a+b+d}=\dfrac{d}{b+c+a}\)

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

Answer 1

[Lê Trung Hiếu]

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

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

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

*Nếu a+b+c+d=0

=> \(\left\{{}\begin{matrix}a+b=-\left(c+d\right)\\b+c=-\left(a+d\right)\end{matrix}\right.\)

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

Nếu a+b+c+d\(\ne\)0

=> a=b=c=d

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

Answer 2

[dmcldt12]

Xét a+b+c+d=0

\(\Rightarrow\)a=-(b+c+d).Thay vào \(\dfrac{a}{b+c+d}\)ta có

\(\dfrac{-\left(b+c+d\right)}{b+c+d}\)=-1.Làm tương tự như thế ta có

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

Xét a+b+c+d\(\ne\)0

Áp dụng tính chất dãy tỉ số bằng nhau ta có

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

=\(\dfrac{a+b+c+d}{2\cdot\left(a+b+c+d\right)}\)=\(\dfrac{1}{2}\)

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

\(\Rightarrow\)2a=b+c+d

\(\Rightarrow\)3a=a+b+c+d\(\left(1\right)\)

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

\(\Rightarrow\)2b= a+c+d

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

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

\(\Rightarrow\)2c=a+b+d

\(\Rightarrow\)3c=a+b+c+d\(\left(3\right)\)

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

\(\Rightarrow\)2d=b+c+a

\(\Rightarrow\)3d=a+b+c+d\(\left(4\right)\)

Từ\(\left(1\right)\),\(\left(2\right)\),\(\left(3\right)\),\(\left(4\right)\)

\(\Rightarrow\)3a=3b=3c=3d

\(\Rightarrow\)a=b=c=d.Khi đó

M=\(\dfrac{a+a}{a+a}+\dfrac{a+a}{a+a}+\dfrac{a+a}{a+a}+\dfrac{a+a}{a+a}\)

=1+1+1+1

=4

Vậy...

Mình trình bày hơi xấu các bạn thông cảm1!