Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
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
Áp dụng dãy tỉ số bằng nhau :
\(\dfrac{a}{b+c+d}=\dfrac{b}{a+c+d}=\dfrac{c}{a+b+d}=\dfrac{d}{b+c+a}=\dfrac{a+b+c+d}{b+c+d+a+c+d+a+b+d+b+c+a}=\dfrac{1}{3}\) \(\Rightarrow3a=b+c+d\left(1\right)\)
\(\Rightarrow3b=c+d+a\left(2\right)\)
\(\Rightarrow3c=a+b+d\left(3\right)\)
\(\Rightarrow3d=b+c+a\left(4\right)\)
Từ \(\left(1\right)+\left(2\right)\Rightarrow3a+3b=b+c+d+c+d+a\)
\(\Rightarrow2a+2b=2c+2d\)
\(\Rightarrow a+b=c+d\)
Từ \(\left(2\right)+\left(3\right)\Rightarrow3b+3c=a+c+d+a+b+c\)
\(\Rightarrow2b+2c=2d+2a\)
\(\Rightarrow b+c=d+a\)
Từ \(\left(1\right)+\left(3\right)\Rightarrow2a+2c=2b+2d\)
\(\Rightarrow a+c=b+d\)
Ta có :
\(b+c=a+d;a+c=b+d\)
\(\Rightarrow b+c+a+c=d+a+b+a\)
\(\Rightarrow a+b+2c=2a+a+d\)
\(\Rightarrow c=d\)
Lại có :
\(b+c=d+a;a+c=b+d\)
\(\Rightarrow b+c+b+d=d+a+a+c\)
\(\Rightarrow2b+c+d=2a+d+c\)
\(\Rightarrow a=b\)
Từ những điều trên ta thấy được :
\(\dfrac{a+b}{c+d}+\dfrac{b+c}{a+d}+\dfrac{c+d}{a+b}+\dfrac{d+a}{b+c}=1+1+1+1=4\)
Nguyễn Thanh Hằng Xét thiếu TH rồi bạn !!!
Ta có :
\(\dfrac{a}{b+c+d}=\dfrac{b}{a+c+d}=\dfrac{c}{a+b+d}=\dfrac{d}{a+b+c}\\ \Rightarrow\dfrac{a}{b+c+d}+1=\dfrac{b}{a+c+d}+1=\dfrac{c}{a+b+d}+1=\dfrac{d}{a+b+c}+1\\ \Rightarrow\dfrac{a+b+c+d}{b+c+d}=\dfrac{a+b+c+d}{a+c+d}=\dfrac{a+b+c+d}{a+b+d}=\dfrac{a+b+c+d}{a+b+c}\)
TH1: Nếu a+b+c+d#0
thì Đỗ Thu Trà giải giống bạn Nguyễn Thanh Hằng
Nếu a+b+c+d=0 =>a+b=-(c+d); b+c=-(a+d);c+d=-(a+b); a+d=-(b+c)
Thế những cái này vao biểu thức M thì M=-4
\(\dfrac{a}{b+c+d}=\dfrac{b}{a+c+d}=\dfrac{c}{a+b+d}=\dfrac{d}{a+b+c}=\dfrac{a+b+c+d}{3\left(a+b+c+d\right)}\dfrac{1}{3}\)(vìa+b+c+d\(\ne\)0)
=>3a=b+c+d: 3b=a+c+d=>3a-3b=b-a
=>3(a-b)=-(a-b)=>4(a-b)=0=>a=b
Tương tự => a=b=c=d=> A=4
Ta có: \(\dfrac{a}{b+c+d}=\dfrac{b}{a+c+d}=\dfrac{c}{a+b+d}=\dfrac{d}{a+b+c}=\dfrac{a+b+c+d}{3\left(a+b+c+d\right)}=\dfrac{1}{3}\)
Ta có: \(\dfrac{a}{b+c+d}=\dfrac{b}{a+c+d}=\dfrac{a+b}{a+b+2\left(c+d\right)}=\dfrac{1}{3}\)
\(\Rightarrow3\left(a+b\right)=\left(a+b\right)+2\left(c+d\right)\)
\(\Rightarrow2\left(a+b\right)=2\left(c+d\right)\)
\(\Rightarrow a+b=c+d\)
\(\Rightarrow\dfrac{a+b}{c+d}=1\)
Tương tự:\(\dfrac{b+c}{a+d}=1;\dfrac{c+d}{a+b}=1;\dfrac{d+a}{b+c}=1\)
Vậy A=4.
Ta có:
\(\dfrac{2a+b+c+d}{a}=\dfrac{a+2b+c+d}{b}=\dfrac{a+b+2c+d}{c}=\dfrac{a+b+c+2d}{d}\)\(\Rightarrow\dfrac{2a+b+c+d}{a}-1=\dfrac{a+2b+c+d}{b}-1=\dfrac{a+b+2c+d}{c}-1=\dfrac{a+b+c+2d}{d}-1\)
\(\dfrac{a+b+c+d}{a}=\dfrac{a+b+c+d}{b}=\dfrac{a+b+c+d}{c}=\dfrac{a+b+c+d}{d}\)
+) Nếu \(a+b+c+d\ne0\) thì từ trên suy ra:\(a=b=c=d\)
\(\Rightarrow M=\dfrac{a+a}{a+a}+\dfrac{a+a}{a+a}+\dfrac{a+a}{a+a}+\dfrac{a+a}{a+a}=1+1+1+1=4\)
+) Nếu \(a+b+c+d=0\)
\(\Rightarrow\left\{{}\begin{matrix}a+b=-\left(c+d\right)\\b+c=-\left(d+a\right)\\c+d=-\left(a+b\right)\\d+a=-\left(b+c\right)\end{matrix}\right.\)
\(\Rightarrow M=\dfrac{-\left(c+d\right)}{c+d}+\dfrac{-\left(d+a\right)}{d+a}+\dfrac{-\left(a+b\right)}{a+b}+\dfrac{-\left(b+c\right)}{b+c}=\left(-1\right)+\left(-1\right)+\left(-1\right)+\left(-1\right)=\left(-4\right)\)
Vậy M = 4 hoặc M = -4
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
tks bạn nhìu nha NGUYỄN HUY TÚ