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Áp dụng tính chất của dãy tỉ số = nhau ta có:
\(\frac{a}{b}=\frac{b}{c}=\frac{c}{d}=\frac{d}{a}=\frac{a+b+c+d}{b+c+d+a}=1\)
=> a = b = c = d
=> \(D=\frac{2a-a}{2a-a}+\frac{2a-a}{2a-a}+\frac{2a-a}{2a-a}+\frac{2a-a}{2a-a}\)
D = 1 + 1 + 1 + 1 = 4
Ta có : 2a + b + c+ d / a - 1 = a + 2b + c + d / b - 1 = a + b + 2c + d / c - 1 = a + b + c +2d / d - 1
=> a + b + c + d / a = a + b + c + d / b = a + b + c + d / c = a + b + c + d / d
Xét 2 trường hợp :
TH1: a + b + c + d = 0
=> a + b = - ( c + d ) ; b + c = - ( a + d ) ; c + d = - ( a + b )
Khi đó M = ( -1 ) . 4 = -4
TH2 : a + b + c + d khác 0
=> a = b = c = d
Khi đó M = 1 . 4 = 4
Vậy M = 4 hoặc M = - 4
Tham khảo nhé
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Ta có: \(\frac{2a+b+c+d}{a}=\frac{a+2b+c+d}{b}=\frac{a+b+2c+d}{c}=\frac{a+b+c+2d}{d}\)
\(=\frac{2a+b+c+d+a+2b+c+d+a+b+2c+d+a+b+c+2d}{a+b+c+d}=4\)
=>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
\(\Rightarrow\frac{a+b}{c+d}=2\)
Tương tự ta có:\(b+\frac{c}{d}+a=2\)
\(c+\frac{d}{a}+b=2\)
\(d+\frac{a}{b}+c=2\)
=>M=2+2+2+2=8
Bạn tham khảo tại đây:
Câu hỏi của Nguyễn Quỳnh Chi - Toán lớp 7 - Học toán với OnlineMath
\(A=\frac{2a+b+c+d}{a}=\frac{a+2b+c+d}{b}=\frac{a+b+2c+d}{c}=\frac{a+b+c+2d}{d}\)
\(\Rightarrow\frac{a+b+c+d}{a}=\frac{a+b+c+d}{b}=\frac{a+b+c+d}{c}=\frac{a+b+c+d}{d}\)
Với a + b + c + d = 0 => a + b = - ( c + d )
=> \(A=\left(-1\right)+\left(-1\right)+\left(-1\right)+\left(-1\right)=-4\)
Với \(a+b+c+d\ne0\) => a = b = c = d
=> \(A=1+1+1+1=4\)
Ta có: \(\frac{2a+b+c+d}{a}=\frac{a+2b+c+d}{b}=\frac{a+b+2c}{c}=\frac{a+b+c+2d}{d}\)
\(\Rightarrow\frac{2a+b+c+d}{a}-1=\frac{a+2b+c+d}{b}-1=\frac{a+b+2c+d}{c}-1=\frac{a+b+c+2d}{d}-1\)
\(\Rightarrow\frac{a+b+c+d}{a}=\frac{a+b+c+d}{b}=\frac{a+b+c+d}{c}=\frac{a+b+c+d}{d}\)(1)
TH1: a + b + c + d =0
=> a + b = -c - d
b + c = - a - d
a + c = -b - d
\(\Rightarrow\frac{a+b}{c+d}+\frac{b+c}{a+d}+\frac{c+a}{b+d}\)
\(=\frac{-c-d}{c+d}+\frac{-a-d}{a+d}+\frac{-b-d}{b+d}\)
\(=\frac{-\left(c+d\right)}{c+d}+\frac{-\left(a+d\right)}{a+d}+\frac{-\left(b+d\right)}{b+d}\)
\(=-1+\left(-1\right)+\left(-1\right)=-3\)
TH2: \(a+b+c+d\ne0\)
Từ (1) => a = b = c =d
\(\Rightarrow\frac{a+b}{c+d}+\frac{b+c}{a+d}+\frac{c+a}{b+d}\)
\(=\frac{a+a}{a+a}+\frac{b+b}{b+b}+\frac{c+c}{c+c}\)
\(=1+1+1=3\)
\(\frac{2a+b+c+d}{a}=\frac{a+2b+c+d}{b}=\frac{a+b+2c+d}{c}=\frac{a+b+c+2d}{d}=\)
\(=\frac{a+b+2c+d+a+b+c+2d}{c+d}=\frac{2\left(a+b\right)}{c+d}+3=\)
Tương tự
\(=\frac{2\left(b+c\right)}{d+a}+3=\)
\(=\frac{2\left(c+d\right)}{a+b}+3=\)
\(=\frac{2\left(d+a\right)}{b+c}+3\)
\(\Rightarrow\frac{2\left(a+b\right)}{c+d}+3=\frac{2\left(b+c\right)}{d+a}+3=\frac{2\left(c+d\right)}{a+b}+3=\frac{2\left(d+a\right)}{b+c}+3\)
\(\Rightarrow\frac{2\left(a+b\right)}{c+d}=\frac{2\left(b+c\right)}{d+a}=\frac{2\left(c+d\right)}{a+b}=\frac{2\left(d+a\right)}{b+c}=\)
\(=\frac{2\left(a+b\right)+2\left(b+c\right)+2\left(c+d\right)+2\left(d+a\right)}{c+d+d+a+a+b+b+c}=\frac{4\left(a+b+c+d\right)}{2\left(a+b+c+d\right)}=2\)
\(\Rightarrow\frac{a+b}{c+d}+\frac{b+c}{d+a}+\frac{c+d}{a+b}+\frac{d+a}{b+c}=1+1+1+1=4\)