Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
Vì a,b,c,d>0 ta áp dụng t/c dãy tỉ số bằng nhau:
`a/(2b)=b/(2c)=c/(2d)=d/(2a)=(a+b+c+d)/(2a+2b+2c+2d)=1/2`
`=>a/(2b)=1/2=>a=b`
Tương tự ta có:`b=c,c=d,d=a`
`=>a=b=c=d`
`=>A=(2011a-2010a)/(a+a)+(2011a-2010a)/(a+a)+(2011a-2010a)/(a+a)+(2011a-2010a)/(a+a)=1/2+1/2+1/2+1/2=2`
Áp dụng tính chất của dãy tỉ số bằng nhau, ta được:
\(\dfrac{a}{2b}=\dfrac{b}{2c}=\dfrac{c}{2d}=\dfrac{d}{2a}=\dfrac{a+b+c+d}{2b+2c+2d+2a}=\dfrac{1}{2}\)
Do đó:
\(\left\{{}\begin{matrix}\dfrac{a}{2b}=\dfrac{1}{2}\\\dfrac{b}{2c}=\dfrac{1}{2}\\\dfrac{c}{2d}=\dfrac{1}{2}\\\dfrac{d}{2a}=\dfrac{1}{2}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}a=b\\b=c\\c=d\\d=a\end{matrix}\right.\Leftrightarrow a=b=c=d\)
Ta có: \(A=\dfrac{2011a-2010b}{c+d}+\dfrac{2011b-2010c}{d+a}+\dfrac{2011c-2010d}{a+b}+\dfrac{2011d-2010a}{b+c}\)
\(=\dfrac{a}{2a}+\dfrac{a}{2a}+\dfrac{a}{2a}+\dfrac{a}{2a}=2\)
áp dụng tính chất dãy tỉ số = nhau ta có
a/b=b/c=c/d=d/a=a=b=c=d/b=c=d=a=1
suy ra; M=2a-a/a+a+2a-a/a+a+2a-a.a+a+2a-a/a+a
=a/2a*4=2
Vậy M=2
\(\frac{a}{b}=\frac{b}{c}=\frac{c}{d}=\frac{d}{a}=\frac{a+b+c+d}{b+c+d+a}=1\)
=>a=b=c=d=>\(a+b=\frac{1}{2}\left(a+b+c+d\right)\)
\(\Rightarrow\frac{2a-b}{c+d}+\frac{2b-c}{d+a}+\frac{2c-d}{a+b}+\frac{2d-a}{b+c}=\frac{2a-b+2b-c+2c-d+2d-a}{a+b}\)
\(=\frac{2\left(a+b+c+d\right)-\left(a+b+c+d\right)}{\frac{1}{2}\left(a+b+c+d\right)}=\frac{a+b+c+d}{\frac{1}{2}\left(a+b+c+d\right)}=\frac{1}{\frac{1}{2}}=2\)
vậy A=2
\(\frac{a}{b}=\frac{b}{c}=\frac{c}{d}=\frac{d}{a}=\frac{a+b+c+d}{b+c+d+a}=1\)
\(\Rightarrow a=b=c=d\)
\(\Rightarrow\frac{2a-b}{c+d}+\frac{2b-c}{d+a}+\frac{2c-d}{a+b}+\frac{2d-a}{b+c}=\frac{2a-a}{a+a}+\frac{2a-a}{a+a}+\frac{2a-a}{a+a}+\frac{2a-a}{a+a}\)
\(\frac{2a-a}{a+a}.4=\frac{a}{2a}.4=\frac{4a}{2a}=2\)
vậy A=2
Bài 1: Đặt \(\dfrac{a}{c}=\dfrac{b}{d}=k\)
\(\Leftrightarrow\left\{{}\begin{matrix}a=ck\\b=dk\end{matrix}\right.\)
\(\dfrac{a}{a+c}=\dfrac{ck}{ck+c}=\dfrac{ck}{c\left(k+1\right)}=\dfrac{k}{k+1}\)
\(\dfrac{b}{b+d}=\dfrac{dk}{dk+d}=\dfrac{k}{k+1}\)
Do đó: \(\dfrac{a}{a+c}=\dfrac{b}{b+d}\)
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: \(a+b+c+d=0\)
\(\Rightarrow\left\{{}\begin{matrix}a+b=-\left(c+d\right)\\b+c=-\left(d+a\right)\\c+d=-\left(a+b\right)\\d+a=-\left(b+c\right)\end{matrix}\right.\)
\(P=\dfrac{a+b}{c+d}+\dfrac{b+c}{d+a}+\dfrac{c+d}{a+b}+\dfrac{d+a}{b+c}\)
\(\Rightarrow P=\dfrac{-\left(c+d\right)}{c+d}+\dfrac{-\left(d+a\right)}{d+a}+\dfrac{-\left(a+b\right)}{a+b}+\dfrac{-\left(b+c\right)}{b+c}\)
\(\Rightarrow P=-1+\left(-1\right)+\left(-1\right)+\left(-1\right)\)
\(\Rightarrow P=-4\)
bn mình nền của bn là nôb team trưởng team là t gaming