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Ta có : \(\frac{a}{2009}=\frac{b}{2011}=\frac{c}{2013}=\frac{a-b}{-2}=\frac{b-c}{-2}=\frac{a-c}{-4}\)
\(=>\frac{\left(a-c\right)^2}{16}=\left(\frac{a-b}{-2}\right).\left(\frac{b-c}{-2}\right)=\frac{\left(a-b\right).\left(b-c\right)}{4}\)
\(=>\frac{\left(a-c\right)^2}{4}=\left(a-b\right).\left(b-c\right)\)
Áp dụng t/c dãy tỉ số bằng nhau,ta có:
\(\frac{a}{2009}=\frac{b}{2011}=\frac{a-b}{2009-2011}=\frac{a-b}{-2}\)
\(\frac{b}{2011}=\frac{c}{2013}=\frac{b-c}{2011-2013}=\frac{b-c}{-2}\)
\(\frac{a}{2009}=\frac{c}{2013}=\frac{a-c}{2009-2013}=\frac{a-c}{-4}\)
=> \(\frac{a-b}{-2}=\frac{b-c}{-2}=\frac{a-c}{-4}\)
=> \(\frac{a-b}{-2}.\frac{b-c}{-2}=\left(\frac{a-c}{4}\right)^2\)
=> \(\frac{\left(a-c\right)^2}{4^2}=\frac{\left(a-b\right)\left(b-c\right)}{4}\)
=> \(\frac{\left(a-c\right)^2}{4}=\left(a-c\right)\left(b-c\right)\)
Ta có : \(\frac{a}{2009}=\frac{b}{2011}=\frac{c}{2013}=\frac{a-b}{-2}=\frac{b-c}{-2}=\frac{a-c}{-4}\)
\(=>\frac{\left(a-c\right)^2}{16}=\left(\frac{a-b}{-2}\right).\left(\frac{b-c}{-2}\right)=\frac{\left(a-b\right).\left(b-c\right)}{4}\)
\(=>\frac{\left(a-c\right)^2}{4}=\left(a-b\right).\left(b-c\right)\)
Đặt a/2019=b/2021=c/2023=k
=>a=2019k; b=2021k; c=2023k
(a-c)^2/4=(2023k-2019k)^2/4=(4k)^2/4=4k^2
(a-b)(b-c)=(2019k-2021k)(2021k-2023k)=4k^2
=>(a-c)^2/4=(a-b)(b-c)
đặt a/2003=b/2005=c/2007=t
=>a=2003t;b=2005t;c=2007t
ta có:\(VT=\frac{\left(a-c\right)^2}{4}=\frac{\left(2003t-2007t\right)^2}{4}=\frac{\left(-4t\right)^2}{4}=\frac{\left(-4\right)^2.t^2}{4}=\frac{16.t^2}{4}=\frac{4.4.t^2}{4}=4t^2\) (1)
\(VP=\left(a-b\right)\left(b-c\right)=\left(2003t-2005t\right)\left(2005t-2007t\right)=\left(-2\right).t.\left(-2\right).t=\left[\left(-2\right).\left(-2\right)\right].t^2=4t^2\left(2\right)\)
từ (1);(2) ta có VT=VP=>đpcm
Ta có:\(\frac{3a+b+c+d}{a}=\frac{a+3b+c+d}{b}=\frac{a+b+3c+d}{c}=\frac{a+b+c+3d}{d}\)
\(\Rightarrow\frac{a+b+c+d}{a}=\frac{a+b+c+d}{b}=\frac{a+b+c+d}{c}=\frac{a+b+c+d}{d}\)
\(\Rightarrow\orbr{\begin{cases}a+b+c+d=0\\a=b=c=d\end{cases}}\)
\(TH1:a+b+c+d=0\Rightarrow\hept{\begin{cases}a+b=-\left(c+d\right)\\b+c=-\left(a+d\right)\end{cases}}\)
\(\Rightarrow Q=\left(\frac{-\left(c+d\right)}{c+d}\right)^2+\left(\frac{-\left(a+d\right)}{a+d}\right)^2+\left(\frac{c+d}{-\left(c+d\right)}\right)^2+\left(\frac{a+d}{-\left(a+d\right)}\right)^2\)
\(\Rightarrow Q=\left(-1\right)^2\cdot4=1\cdot4=4\)
\(TH2:a=b=c=d\)
\(\Rightarrow Q=\left(\frac{a+a}{a+a}\right)^2+\left(\frac{a+a}{a+a}\right)^2+\left(\frac{a+a}{a+a}\right)^2+\left(\frac{a+a}{a+a}\right)^2=1^2\cdot4=1\cdot4=4\)
Vậy Q=4
Ta có :
\(\frac{a+b-b-c}{2018-2019}=\frac{a-c}{-1}\)
\(\frac{b+c-c-a}{2019-2020}=\frac{b-a}{-1}\)
\(\frac{b-c}{2018-2020}=\frac{b-c}{-2}\)
Đặt \(\frac{a-c}{-1}=\frac{b-a}{-1}=\frac{b-c}{-2}=k\left(k\ne0\right)\)
\(\Rightarrow\hept{\begin{cases}\frac{a-c}{-1}=k\\\frac{b-a}{-1}=k\\\frac{b-c}{-2}=k\end{cases}\Rightarrow\hept{\begin{cases}a-c=-k\\b-a=-k\\b-c=k.\left(-2\right)\end{cases}}}\)
\(\Rightarrowđpcm\)