Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
\(\frac{a}{b}=\frac{b}{c}=\frac{c}{d}=\frac{d}{a}=\frac{a+b+c+d}{b+c+d+a}=1\) (vì a + b + c + d khác 0) nên a = b = c = d
\(\Rightarrow\frac{2a-b}{c+d}+\frac{2b-c}{d+a}+\frac{2c-d}{a+b}+\frac{2d-a}{b+c}=\frac{2a-a}{a+a}+\frac{2a-a}{a+a}+\frac{2a-a}{a+a}+\frac{2a-a}{a+a}\)
\(=\frac{1}{2}.4=2\)
Áp dụng tính chất của dãy tỉ số bằng nhau, ta được:
\(\frac{a}{b}=\frac{b}{c}=\frac{c}{d}=\frac{d}{a}=\frac{a+b+c+d}{a+b+c+d}=1\)
\(\Rightarrow\frac{a}{b}=\frac{b}{c}=\frac{c}{d}=\frac{d}{a}=1\)
\(\Rightarrow a=b=c=d\)
Khí đó:
\(M=\frac{a+b}{c+d}+\frac{b+c}{d+a}+\frac{c+d}{a+b}+\frac{d+a}{b+c}\)
\(M=\frac{a+a}{a+a}+\frac{a+a}{a+a}+\frac{a+a}{a+a}+\frac{a+a}{a+a}=4\)
Vậy M = 4
Giải: Áp dụng t/c của dãy tỉ số bằng nhau, ta có:
\(\frac{a}{b}=\frac{b}{c}=\frac{c}{d}=\frac{d}{a}=\frac{a+b+c+d}{b+c+d+a}=1\) (vì a + b + c + d\(\ne\)0)
=> \(\frac{a}{b}=1\)=> a = b
\(\frac{b}{c}=1\) => b = c
\(\frac{c}{d}=1\) => c = d
\(\frac{d}{a}=1\) => d = a
=> a = b = c = d
Khi đó, ta có: \(\frac{2a-b}{c+d}+\frac{2b-c}{a+d}+\frac{2c-d}{a+b}+\frac{2d-a}{b+c}\)
hay \(\frac{2a-a}{a+a}+\frac{2a-a}{a+a}+\frac{2a-a}{a+a}+\frac{2a-a}{a+a}\)
\(=\frac{a}{2a}+\frac{a}{2a}+\frac{a}{2a}+\frac{a}{2a}\)
= \(\frac{1}{2}+\frac{1}{2}+\frac{1}{2}+\frac{1}{2}\)
= \(\frac{1}{2}.4=2\)
Vì \(\frac{a}{b+c+d}\)= \(\frac{b}{a+c+d}\)= \(\frac{c}{a+b+d}\)= \(\frac{d}{a+b+c}\)nên
\(\frac{a}{b+c+d}\)+1 = \(\frac{b}{a+c+d}\)+1 = \(\frac{c}{a+b+d}\)+1 = \(\frac{d}{a+b+c}\) +1
hay\(\frac{a+b+c+d}{b+c+d}\) = \(\frac{a+b+c+d}{a+c+d}\)= \(\frac{a+b+c+d}{a+b+d}\)= \(\frac{a+b+c+d}{a+b+c}\)
Mà a + b + c + d \(\ne\)0 \(\Rightarrow\) \(b+c+d=a+c+d=a+b+d=a+b+c\)
\(\Rightarrow\) \(a=b=c=d\)
\(\Rightarrow\) \(M=4\)
Ta có:\(\frac{a}{b+c+d}=\frac{b}{c+d+a}=\frac{c}{a+b+d}=\frac{d}{a+b+c}\)
\(\Rightarrow\frac{b+c+d}{a}=\frac{c+d+a}{b}=\frac{a+b+d}{c}=\frac{a+b+c}{d}\)
\(\Rightarrow\frac{b+c+d}{a}+1=\frac{c+d+a}{b}+1=\frac{a+b+d}{c}+1=\frac{a+b+c}{d}+1\)
\(\Rightarrow\frac{a+b+c+d}{a}=\frac{a+b+c+d}{b}=\frac{a+b+c+d}{c}=\frac{a+b+c+d}{d}\)
Vì a+b+c+d\(\ne\)0=>a=b=c=d
\(\Rightarrow A=\frac{a+c}{b+d}+\frac{a+b}{c+d}+\frac{a+c}{b+d}+\frac{b+c}{a+d}=\frac{a+a}{a+a}+\frac{a+a}{a+a}+\frac{a+a}{a+a}+\frac{a+a}{a+a}=1+1+1+1=4\)
còn ai nữa à =='
đk a,b,c,d khác 0
áp dugnj tc dãy tỉ số = nhau \(\frac{2a+b+c+d}{a}=\frac{a+2b+c+d}{b}=\frac{a+b+2c+d}{c}=\frac{a+b+c+2d}{d}\)
\(=\frac{2a+b+c+d+a+2b+c+d+a+b+2c+d+a+b+c+2d}{a+b+c+d}=\frac{5\left(a+b+c+d\right)}{a+b+c+d}\)
+> nếu a+b+c+d =0\(\Rightarrow\hept{\begin{cases}a+b=-\left(c+d\right)\\b+c=-\left(d+a\right)\\c+d=-\left(a+b\right)\end{cases}\hept{\begin{cases}d+a=-\left(b+c\right)\\\end{cases}}}\)\(\Rightarrow M=-4\)
+> a+b+c+d khác 0 \(\Rightarrow\frac{2a+b+c+d}{a}=5\Rightarrow b+c+d=3a\)
Tương tự ta có \(\hept{\begin{cases}a+b+c=3d\\a+c+d=3b\\a+b+d=3c\end{cases}}\)\(\Rightarrow a=b=c=d\)
Khi đó M=4
Vậy M=4 hoặc M=-4
\(\Leftrightarrow\frac{a}{b+c+d}+1=\frac{b}{c+d+a}+1=\frac{c}{a+b+d}+1=\frac{d}{a+b+c}+1\)
\(\Leftrightarrow\frac{a}{a+b+c+d}=\frac{b}{a+b+c+d}=\frac{c}{a+b+c+d}=\frac{d}{a+b+c+d}\)
\(\Rightarrow a=b=c=d\) Thay vào A ta được :
\(A=\frac{a+a}{a+a}+\frac{a+a}{a+a}+\frac{a+a}{a+a}+\frac{a+a}{a+a}=1+1+1+1=4\)
Cảm ơn bạn nhiều