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\(\dfrac{a}{b+c+d}=\dfrac{b}{c+a+d}=\dfrac{c}{a+b+d}=\dfrac{d}{a+b+c}\)
+)Xét a+b+c+d=0 thì a+d=-c-d
b+c=-d-a
c+d=-b-a
d+a=-b-c
Do đó:
\(P=\dfrac{-c-d}{c+d}+\dfrac{-a-b}{a+b}+\dfrac{-b-c}{b+c}+\dfrac{-d-a}{a+d}\\ =-1+-1+-1+-1=-4\)
+)Xét a+b+c+d khác 0
áp dụng t/c của dãy tỉ số bằng nhau, ta có:
\(\dfrac{a}{b+c+d}=\dfrac{b}{a+c+d}=\dfrac{c}{a+b+d}=\dfrac{d}{a+c+d}=\dfrac{a+b+c+d}{3\left(a+b+c+d\right)}=\dfrac{1}{3}\)
=>\(a=\dfrac{1}{3}\left(d+b+c\right)\)
\(b=\dfrac{1}{3}\left(a+c+d\right)\)
\(c=\dfrac{1}{3}\left(a+b+d\right)\)
\(d=\dfrac{1}{3}\left(a+b+c\right)\)
Bạn thay vào r tính
Ta có : \(\dfrac{a}{b+c+d}=\dfrac{b}{c+d+a}=\dfrac{c}{d+a+b}=\dfrac{d}{a+b+c}\)
\(\Rightarrow\)\(\dfrac{a}{b+c+d}+1=\dfrac{b}{c+d+a}+1=\dfrac{c}{d+a+b}+1=\dfrac{d}{a+b+c}+1\)
\(\Rightarrow\)\(\dfrac{a+b+c+d}{b+c+d}=\dfrac{b+c+d+a}{c+d+a}=\dfrac{c+d+a+b}{d+a+b}=\dfrac{d+a+b+c}{a+b+c}\)
TH1 : \(a+b+c+d\ne0\)\(\Rightarrow\) \(a=b=c=d\)
\(\Rightarrow\) P= \(\dfrac{a+b}{c+d}+\dfrac{b+c}{d+a}+\dfrac{c+d}{b+a}+\dfrac{d+a}{b+c}=1+1+1+1=4\)
TH2 : \(a+b+c+d=0\Rightarrow\left\{{}\begin{matrix}a+b=-\left(c+d\right)\\b+c=-\left(a+d\right)\end{matrix}\right.\)
\(\Rightarrow P=\dfrac{a+b}{c+d}+\dfrac{b+c}{d+a}+\dfrac{c+d}{b+a}+\dfrac{d+a}{b+c}=-1+\left(-1\right)+\left(-1\right)+\left(-1\right)=-4\)
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
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!
Giải:
Ta có:
\(\dfrac{a}{b+c+d}=\dfrac{b}{a+c+d}=\dfrac{c}{a+b+d}=\dfrac{d}{b+c+a}\)
Áp dụng tính chất của dãy tỉ số bằng nhau, ta có:
\(\dfrac{a}{b+c+d}=\dfrac{b}{a+c+d}=\dfrac{c}{a+b+d}=\dfrac{d}{b+c+a}=\dfrac{a+b+c+d}{3\left(a+b+c+d\right)}=\dfrac{1}{3}\)
\(\Leftrightarrow\left\{{}\begin{matrix}\dfrac{a}{b+c+d}=\dfrac{1}{3}\\\dfrac{b}{a+c+d}=\dfrac{1}{3}\\\dfrac{c}{a+b+d}=\dfrac{1}{3}\\\dfrac{d}{b+c+a}=\dfrac{1}{3}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}3a=b+c+d\\3b=a+c+d\\3c=a+b+d\\3d=b+c+a\end{matrix}\right.\Leftrightarrow a=b=c=d\)
\(\Leftrightarrow M=\dfrac{a+b}{c+d}+\dfrac{b+c}{a+d}+\dfrac{c+d}{a+b}+\dfrac{d+a}{b+c}\)
\(\Leftrightarrow M=1+1+1+1=4\)
Vậy \(M=4\).
Chúc bạn học tốt!
Ta có
\(\dfrac{2a+b+c+d}{a}=\dfrac{a+2b+c+d}{b}=\dfrac{a+b+2c+d}{c}=\dfrac{a+b+c+2d}{d}\)
\(\Rightarrow\dfrac{2a+b+c+d}{a}-1=\dfrac{a+2b+c+d}{b}-1=\dfrac{a+b+2c+d}{c}-1=\dfrac{a+b+c+2d}{d}-1\)
\(\Rightarrow\dfrac{a+b+c+d}{a}=\dfrac{a+b+c+d}{b}=\dfrac{a+b+c+d}{c}=\dfrac{a+b+c+d}{d}\)
Trường hợp thứ nhất \(a+b+c+d\ne0\Rightarrow a=b=c=d\)
\(\Rightarrow M=\dfrac{a+b}{c+d}+\dfrac{b+c}{d+a}+\dfrac{c+d}{a+b}+\dfrac{d+a}{b+c}\)
\(\Rightarrow M=1+1+1+1\)
\(\Rightarrow M=4\)
Trường hợp thứ hai\(a+b+c+d=0\Rightarrow\left\{{}\begin{matrix}a+b=-\left(c+d\right)\\b+c=-\left(d+a\right)\\c+d=-\left(a+b\right)\\d+a=-\left(b+c\right)\end{matrix}\right.\)
\(\Rightarrow M=\dfrac{a+b}{c+d}+\dfrac{b+c}{d+a}+\dfrac{c+d}{a+b}+\dfrac{d+a}{b+c}\)
\(\Rightarrow M=\left(-1\right)+\left(-1\right)+\left(-1\right)+\left(-1\right)\)
\(\Rightarrow M=-4\)
Vậy \(M\in\left\{4;-4\right\}\)
Áp dụng tính chất dãy tỉ số bằng nhau ta có:
\(\dfrac{2a+b+c+d}{a}=\dfrac{a+2b+c+d}{b}=\dfrac{a+b+2c+d}{c}=\dfrac{a+b+c+2d}{d}=\dfrac{2a+b+c+d-a-2b-c-d}{a-b}=1\)
\(\Rightarrow\left\{{}\begin{matrix}-a=b+c+d\\-b=a+c+d\\-c=b+c+d\\-d=a+b+c\end{matrix}\right.\Rightarrow a=b=c=d\)
\(M=\dfrac{a+b}{c+d}+\dfrac{b+c}{a+d}+\dfrac{c+d}{a+b}+\dfrac{a+d}{b+c}\)
\(\Rightarrow M=\dfrac{a+a}{a+a}+\dfrac{a+a}{a+a}+\dfrac{a+a}{a+a}+\dfrac{a+a}{a+a}\)
\(\Rightarrow M=1+1+1+1\)
\(\Rightarrow M=4\)
Vậy .......
Chúc bạn học tốt!
Thiếu trường hợp r bn khi a + b + c + d = 0 thì M = -4
bn phải xét a + b + c + d = 0 và a + b + c + d ≠ 0 khi đó mới đc dùng tính chất dãy tỉ số bằng nhau nha
khi a + b + c + d = 0
⇒ a + b = -(c + d)
a + d = -(b + c)
\(\Rightarrow M=\dfrac{a+b}{-\left(a+b\right)}+\dfrac{-\left(a+d\right)}{a+d}+\dfrac{-\left(a+b\right)}{a+b}+\dfrac{a+d}{-\left(a+d\right)}\)\(\Rightarrow M=-4\)
Theo đề bài, ta có:
\(\dfrac{2a+b+c+d}{a}=\dfrac{a+2b+c+d}{b}=\dfrac{a+b+2c+d}{c}=\dfrac{a+b+c+2d}{d}\)
\(\dfrac{2a+b+c+d}{a}-1=\dfrac{a+2b+c+d}{b}-1=\dfrac{a+b+2c+d}{c}-1=\dfrac{a+b+c+2d}{d}-1\)
\(\dfrac{a+b+c+d}{a}=\dfrac{a+b+c+d}{b}=\dfrac{a+b+c+d}{c}=\dfrac{a+b+c+d}{d}\) vì a,b,c,d khác 0
\(\Rightarrow a=b=c=d\)
\(\Rightarrow M=1+1+1+1=4\)
TH1:
\(\dfrac{a}{b+c+d}=\dfrac{b}{c+d+a}=\dfrac{c}{d+a+b}=\dfrac{d}{a+b+c}=\dfrac{a+b+c+d}{3\left(a+b+c+d\right)}=\dfrac{1}{3}\)
\(\Rightarrow\left\{{}\begin{matrix}3a=b+c+d\\3b=a+c+d\\3c=a+b+d\\3d=a+b+c\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}3\left(a-b\right)=b-a\\3\left(b-c\right)=c-b\\3\left(c-d\right)=d-c\\3\left(d-a\right)=a-d\end{matrix}\right.\) \(\Rightarrow a=b=c=d\)
\(\Rightarrow P=\dfrac{2a}{2a}+\dfrac{2a}{2a}+\dfrac{2a}{2a}+\dfrac{2a}{2a}=1+1+1+1=4\)
TH2: \(\dfrac{a}{b+c+d}=\dfrac{b}{a+c+d}=\dfrac{c}{a+b+d}=\dfrac{d}{a+b+c}=\dfrac{a-b}{a-a}=-1\)
\(\Rightarrow-a=b+c+d\Rightarrow a+b+c+d=0\)
\(\Rightarrow\left\{{}\begin{matrix}a+b=-\left(c+d\right)\\a+c=-\left(b+d\right)\\a+d=-\left(b+c\right)\end{matrix}\right.\)
\(\Rightarrow P=\dfrac{-\left(c+d\right)}{c+d}+\dfrac{b+c}{-\left(b+c\right)}+\dfrac{c+d}{-\left(c+d\right)}+\dfrac{-\left(b+c\right)}{b+c}=-1+-1+-1+-1=-4\)
Vậy \(\left[{}\begin{matrix}P=4\\P=-4\end{matrix}\right.\)
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