Cho : \(\dfrac{a}{b+c+d}=\dfrac{b}{a+c+d}=\dfrac{c}{a+b+d}=\dfrac{d}{b+c+a}\)
Tính : M = \(\dfrac{a+b}{c+d}+\dfrac{b+c}{a+d}+\dfrac{c+d}{a+b}+\dfrac{d+a}{b+c}\)
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5 tháng 8 2021
\(TH1:a+b+c+d\ne0\)
\(\dfrac{2a+b+c+d}{a}=\dfrac{a+2b+c+d}{b}=\dfrac{a+b+2c+d}{c}=\dfrac{a+b+c+2d}{d}\)
\(\Rightarrow\dfrac{2a+b+c+d}{a}-1=\dfrac{a+2b+c+d}{b}-1=\dfrac{a+b+2c+d}{c}-1=\dfrac{a+b+c+2d}{d}-1\)
\(\Rightarrow\dfrac{a+b+c+d}{a}=\dfrac{a+b+c+d}{b}=\dfrac{a+b+c+d}{c}=\dfrac{a+b+c+d}{d}\)
\(\Rightarrow a=b=c=d\)
\(M=\dfrac{a+b}{c+d}+\dfrac{b+c}{d+a}+\dfrac{c+d}{a+b}+\dfrac{a+d}{b+c}\)
\(=1+1+1+1\)
\(=4\)
\(TH2:a+b+c+d=0\)
\(\Rightarrow\left\{{}\begin{matrix}a+b=-\left(c+d\right)\\b+c=-\left(d+a\right)\\c+d=-\left(a+b\right)\\d+a=-\left(b+c\right)\end{matrix}\right.\)
\(M=\dfrac{a+b}{c+d}+\dfrac{b+c}{d+a}+\dfrac{c+d}{a+b}+\dfrac{a+d}{b+c}\)
\(=-\dfrac{c+d}{c+d}-\dfrac{d+a}{d+a}-\dfrac{a+b}{a+b}-\dfrac{b+c}{b+c}\)
\(=-1-1-1-1\)
\(=-4\)
MH
4 tháng 1 2022
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
\(M=\dfrac{a+b}{a+b}=\dfrac{b+c}{b+c}=\dfrac{c+d}{c+d}=\dfrac{d+a}{d+a}=1\)
14 tháng 11 2021
\(\dfrac{a}{b+c+d}=\dfrac{b}{a+c+d}=\dfrac{c}{a+b+d}=\dfrac{d}{a+b+c}=\dfrac{a+b+c+d}{3\left(a+b+c+d\right)}=\dfrac{1}{3}\\ \Rightarrow\left\{{}\begin{matrix}b+c+d=3a\\a+c+d=3b\\a+b+d=3c\\a+b+c=3d\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}a+b+c+d=2a\\a+b+c+d=2b\\a+b+c+d=2c\\a+b+c+d=2d\end{matrix}\right.\\ \Rightarrow2a=2b=2c=2d\\ \Rightarrow a=b=c=d\\ \Rightarrow A=\dfrac{a+a}{a+a}+\dfrac{a+a}{a+a}+\dfrac{a+a}{a+a}+\dfrac{a+a}{a+a}=1+1+1+1=4\)
3 tháng 12 2021
TH1: \(a+b+c+d\ne0\)
\(\dfrac{2a+b+c+d}{a}=\dfrac{a+2b+c+d}{b}=\dfrac{a+b+2c+d}{c}=\dfrac{a+b+c+2d}{d}\)
\(\Rightarrow\dfrac{2a+b+c+d}{a}-1=\dfrac{a+2b+c+d}{b}-1=\dfrac{a+b+2c+d}{c}-1=\dfrac{a+b+c+2d}{d}-1\)
\(\Rightarrow\dfrac{a+b+c+d}{a}=\dfrac{a+b+c+d}{b}=\dfrac{a+b+c+d}{c}=\dfrac{a+b+c+2d}{d}\)
\(\Rightarrow a=b=c=d\)
\(P=\dfrac{a+b}{c+d}+\dfrac{b+c}{d+a}+\dfrac{c+d}{a+b}+\dfrac{d+a}{b+c}\)
\(\Rightarrow P=1+1+1+1\)
\(\Rightarrow P=4\)
TH2: \(a+b+c+d=0\)
\(\Rightarrow\left\{{}\begin{matrix}a+b=-\left(c+d\right)\\b+c=-\left(d+a\right)\\c+d=-\left(a+b\right)\\d+a=-\left(b+c\right)\end{matrix}\right.\)
\(P=\dfrac{a+b}{c+d}+\dfrac{b+c}{d+a}+\dfrac{c+d}{a+b}+\dfrac{d+a}{b+c}\)
\(\Rightarrow P=\dfrac{-\left(c+d\right)}{c+d}+\dfrac{-\left(d+a\right)}{d+a}+\dfrac{-\left(a+b\right)}{a+b}+\dfrac{-\left(b+c\right)}{b+c}\)
\(\Rightarrow P=-1+\left(-1\right)+\left(-1\right)+\left(-1\right)\)
\(\Rightarrow P=-4\)
4 tháng 12 2021
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S
10 tháng 10 2017
\(\frac{2a+b+c+d}{a}=\frac{a+2b+c+d}{b}=\frac{a+b+2c+d}{c}=\frac{a+b+c+2d}{d}\)
\(\Rightarrow\frac{2a+b+c+d}{a}-1=\frac{a+2b+c+d}{b}-1=\frac{a+b+2c+d}{c}-1=\frac{a+b+c+2d}{d}-1\)
\(\Rightarrow\frac{a+b+c+d}{a}=\frac{a+b+c+d}{b}=\frac{a+b+c+d}{c}=\frac{a+b+c+d}{d}\)
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
30 tháng 11 2021
\(\dfrac{a}{b+c+d}=\dfrac{b}{a+c+d}=\dfrac{c}{a+b+d}=\dfrac{d}{a+b+c}\\ \Rightarrow\dfrac{b+c+d}{a}=\dfrac{a+c+d}{b}=\dfrac{a+b+d}{c}=\dfrac{a+b+c}{d}=\dfrac{3\left(a+b+c+d\right)}{a+b+c+d}=3\\ \Rightarrow\left\{{}\begin{matrix}b+d+c=3a\\a+c+d=3b\\a+b+d=3c\\a+b+c=3d\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}a+b+c+d=4a\\a+b+c+d=4b\\a+b+c+d=4c\\a+b+c+d=4d\end{matrix}\right.\\ \Rightarrow4a=4b=4c=4d\Rightarrow a=b=c=d\\ \Rightarrow P=\dfrac{a+a}{a+a}+\dfrac{a+a}{a+a}+\dfrac{a+a}{a+a}+\dfrac{a+a}{a+a}=1+1+1+1=4\)
5 tháng 4 2017
Giải:
Ta có: \(\dfrac{2012a+b+c+d}{a}=\dfrac{a+2012b+c+d}{b}=\dfrac{a+b+2012c+d}{c}\)
\(=\dfrac{a+b+c+2012d}{d}\)
\(\Rightarrow\dfrac{2012a+b+c+d}{a}-2011=\dfrac{a+2012b+c+d}{b}-2011\)
\(=\dfrac{a+b+2012c+d}{c}-2011=\dfrac{a+b+c+2012d}{d}-2011\)
\(\Rightarrow\dfrac{a+b+c+d}{a}=\dfrac{a+b+c+d}{b}=\dfrac{a+b+c+d}{c}=\dfrac{a+b+c+d}{d}\)
+) Xét \(a+b+c+d=0\) ta có:
\(\left\{{}\begin{matrix}a+b=-\left(c+d\right)\\b+c=-\left(a+d\right)\\c+d=-\left(a+b\right)\\d+a=-\left(b+c\right)\end{matrix}\right.\)
\(M=\dfrac{a+b}{c+d}=\dfrac{b+c}{d+a}=\dfrac{c+d}{a+b}=\dfrac{d+a}{b+c}\)
\(\Rightarrow M=\dfrac{-\left(c+d\right)}{c+d}=\dfrac{-\left(a+d\right)}{a+d}=\dfrac{-\left(a+b\right)}{a+b}=\dfrac{-\left(b+c\right)}{b+c}=-1\)
+) Xét \(a+b+c+d\ne0\)
\(\Rightarrow a=b=c=d\)
\(M=\dfrac{a+b}{c+d}=\dfrac{b+c}{d+a}=\dfrac{c+d}{a+b}=\dfrac{d+a}{b+c}\)
\(\Rightarrow M=\dfrac{2a}{2a}=\dfrac{2a}{2a}=\dfrac{2a}{2a}=\dfrac{2a}{2a}=1\)
Vậy nếu \(a+b+c+d=0\) thì M = -1
nếu \(a+b+c+d\ne0\) thì M = 1