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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
\(M=\dfrac{a+b}{a+b}=\dfrac{b+c}{b+c}=\dfrac{c+d}{c+d}=\dfrac{d+a}{d+a}=1\)
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)
\(\Rightarrow M=\dfrac{a+b}{-\left(a+b\right)}+\dfrac{-\left(a+d\right)}{a+d}+\dfrac{-\left(a+b\right)}{a+b}+\dfrac{a+d}{-\left(a+d\right)}\)\(\Rightarrow M=-4\)
\(TH1:a+b+c+d\ne0\)
\(\dfrac{2a+b+c+d}{a}=\dfrac{a+2b+c+d}{b}=\dfrac{a+b+2c+d}{c}=\dfrac{a+b+c+2d}{d}\)
\(\Rightarrow\dfrac{2a+b+c+d}{a}-1=\dfrac{a+2b+c+d}{b}-1=\dfrac{a+b+2c+d}{c}-1=\dfrac{a+b+c+2d}{d}-1\)
\(\Rightarrow\dfrac{a+b+c+d}{a}=\dfrac{a+b+c+d}{b}=\dfrac{a+b+c+d}{c}=\dfrac{a+b+c+d}{d}\)
\(\Rightarrow a=b=c=d\)
\(M=\dfrac{a+b}{c+d}+\dfrac{b+c}{d+a}+\dfrac{c+d}{a+b}+\dfrac{a+d}{b+c}\)
\(=1+1+1+1\)
\(=4\)
\(TH2:a+b+c+d=0\)
\(\Rightarrow\left\{{}\begin{matrix}a+b=-\left(c+d\right)\\b+c=-\left(d+a\right)\\c+d=-\left(a+b\right)\\d+a=-\left(b+c\right)\end{matrix}\right.\)
\(M=\dfrac{a+b}{c+d}+\dfrac{b+c}{d+a}+\dfrac{c+d}{a+b}+\dfrac{a+d}{b+c}\)
\(=-\dfrac{c+d}{c+d}-\dfrac{d+a}{d+a}-\dfrac{a+b}{a+b}-\dfrac{b+c}{b+c}\)
\(=-1-1-1-1\)
\(=-4\)
\(\dfrac{a}{b+c+d}=\dfrac{b}{a+c+d}=\dfrac{c}{a+b+d}=\dfrac{d}{a+b+c}=\dfrac{a+b+c+d}{3\left(a+b+c+d\right)}=\dfrac{1}{3}\\ \Rightarrow\left\{{}\begin{matrix}b+c+d=3a\\a+c+d=3b\\a+b+d=3c\\a+b+c=3d\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}a+b+c+d=2a\\a+b+c+d=2b\\a+b+c+d=2c\\a+b+c+d=2d\end{matrix}\right.\\ \Rightarrow2a=2b=2c=2d\\ \Rightarrow a=b=c=d\\ \Rightarrow A=\dfrac{a+a}{a+a}+\dfrac{a+a}{a+a}+\dfrac{a+a}{a+a}+\dfrac{a+a}{a+a}=1+1+1+1=4\)
có dãy tỉ số bằng nhau đó thì ta cộng vào rồi rút gọn thì được kết quả là \(\dfrac{2015}{2011}\) nó sẽ bằng với từng biểu thức đó.
Mẫu sẽ cố 2011=2011a; 2011=2011b; 2011=2011c; 2011=2011d
=> a = b = c = d = 1
=> M = 4
\(\frac{2a+b+c+d}{a}=\frac{a+2b+c+d}{b}=\frac{a+b+2c+d}{c}=\frac{a+b+c+2d}{d}\)
\(\Rightarrow\frac{2a+b+c+d}{a}-1=\frac{a+2b+c+d}{b}-1=\frac{a+b+2c+d}{c}-1=\frac{a+b+c+2d}{d}-1\)
\(\Rightarrow\frac{a+b+c+d}{a}=\frac{a+b+c+d}{b}=\frac{a+b+c+d}{c}=\frac{a+b+c+d}{d}\)
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
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...
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