Cho \(\dfrac{2010c-2011b}{2009}\)=\(\dfrac{2011a-2009c}{2010}\)=\(\dfrac{2009b-2010a}{2011}\) CMR \(\dfrac{a}{2009}\)=\(\dfrac{b}{2010}\)=\(\dfrac{c}{2011}\)
K
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KT
Cho: (2010c-2011b)/2009= (2011a-2009c)/2010= (2009b-2010a)/2011
Chứng minh rằng: a/2009=b/2010=c/2011
0
17 tháng 1 2023
A = \(\dfrac{2008}{2009+2010+2011}+\dfrac{2009}{2009+2010+2011}+\dfrac{2010}{2009+2010+2011}\)
Ta có:
\(\dfrac{2008}{2009}>\dfrac{2008}{2009+2010+2011}\)
\(\dfrac{2009}{2010}>\dfrac{2009}{2009+2010+2011}\)
\(\dfrac{2010}{2011}>\dfrac{2010}{2009+2010+2011}\)
Từ 3 điều trên suy ra : A < B
13 tháng 2 2019
=\(\dfrac{1}{2009.\left(\dfrac{1}{2009}+\dfrac{1}{2011}+\dfrac{1}{2010}\right)}+\dfrac{1}{2010.\left(\dfrac{1}{2010}+\dfrac{1}{2009}+\dfrac{1}{2011}\right)}+\dfrac{1}{2011.\left(\dfrac{1}{2011}+\dfrac{1}{2009}+\dfrac{1}{2010}\right)}\)\(=\dfrac{1}{2009}:\left(\dfrac{1}{2009}+\dfrac{1}{2010}+\dfrac{1}{2011}\right)+\dfrac{1}{2010}:\left(\dfrac{1}{2009}+\dfrac{1}{2010}+\dfrac{1}{2011}\right)+\dfrac{1}{2011}:\left(\dfrac{1}{2009}+\dfrac{1}{2010}+\dfrac{1}{2011}\right)\)
\(=\left(\dfrac{1}{2009}+\dfrac{1}{2010}+\dfrac{1}{2011}\right):\left(\dfrac{1}{2009}+\dfrac{1}{2010}+\dfrac{1}{2011}\right)=1\)
S
26 tháng 2 2018
\(B=\dfrac{2008+2009+2010}{2009+2010+2011}=\dfrac{2008}{2009+2010+2011}+\dfrac{2009}{2009+2010+2011}+\dfrac{2010}{2009+2010+2011}\)Ta có : \(\dfrac{2008}{2009}>\dfrac{2008}{2009+2010+2011}\)
\(\dfrac{2009}{2010}>\dfrac{2009}{2009+2010+2011}\)
\(\dfrac{2010}{2011}>\dfrac{2010}{2009+2010+2011}\)\(=>\dfrac{2008}{2009}+\dfrac{2009}{2010}+\dfrac{2010}{2011}>\dfrac{2008+2009+2010}{2009+2010+2011}\)
Hay A > B
26 tháng 2 2018
Ta có :
\(B=\frac{2008+2009+2010}{2009+2010+2011}=\frac{2008}{2009+2010+2011}+\frac{2009}{2009+2010+2011}+\frac{2010}{2009+2010+2011}\)
Vì :
\(\frac{2008}{2009}>\frac{2008}{2009+2010+2011}\)
\(\frac{2009}{2010}>\frac{2009}{2009+2010+2011}\)
\(\frac{2010}{2011}>\frac{2010}{2009+2010+2011}\)
Nên \(\frac{2008}{2009}+\frac{2009}{2010}+\frac{2010}{2011}>\frac{2008}{2009+2010+2011}+\frac{2009}{2009+2010+2011}+\frac{2010}{2009+2010+2011}\)
\(\Rightarrow\)\(A>B\)
Vậy \(A>B\)
26 tháng 2 2018
Ta có: \(B=\frac{2008+2009+2010}{2009+2010+2011}\)
\(=\frac{2008}{2009+2010+2011}+\frac{2009}{2009+2010+2011}+\frac{2010}{2009+2010+2011}\)
Vì \(\frac{2008}{2009}>\frac{2008}{2009+2010+2011}\)
\(\frac{2009}{2010}>\frac{2009}{2009+2010+2011}\)
\(\frac{2010}{2011}>\frac{2010}{2009+2010+2011}\)
nên \(\frac{2008}{2009}+\frac{2009}{2010}+\frac{2010}{2011}>\frac{2008+2009+2010}{2009+2010+2011}\)
hay A > B
Vậy A > B
18 tháng 1 2022
Từ \(\dfrac{a}{2b}=\dfrac{b}{2c}=\dfrac{c}{2d}=\dfrac{d}{2a}\Rightarrow\dfrac{1}{2}.\dfrac{a}{b}=\dfrac{1}{2}.\dfrac{b}{c}=\dfrac{1}{2}.\dfrac{c}{d}=\dfrac{1}{2}.\dfrac{d}{a}\)
⇒ \(\dfrac{a}{b}=\dfrac{b}{c}=\dfrac{c}{d}=\dfrac{d}{a}=\dfrac{a+b+c+d}{b+c+d+a}=1\)
⇒ \(a=b=c=d\)
Thay b = a ; c = a ; d = a vào biểu thức A ta có:
\(A=\dfrac{2011a-2010a}{2a}+\dfrac{2011a-2010a}{2a}+\dfrac{2011a-2010a}{2a}+\dfrac{2011a-2010a}{2a}\)
\(A=\dfrac{a}{2a}+\dfrac{a}{2a}+\dfrac{a}{2a}+\dfrac{a}{2a}\)
\(A=\dfrac{1}{2}.4=2\)
Vậy A = 2
D
18 tháng 1 2022
\(\dfrac{a}{2b}=\dfrac{b}{2c}=\dfrac{c}{2d}=\dfrac{d}{2a}\)
Áp dụng tính chất của dãy tỉ số bằng nhau, ta có:
\(\dfrac{a}{2b}=\dfrac{b}{2c}=\dfrac{c}{2d}=\dfrac{d}{2a}=\dfrac{a+b+c+d}{2a+2b+2c+2d}=\dfrac{1}{2}\)
=>\(\dfrac{a}{2b}=\dfrac{1}{2}\)=>2a=2b =>a=b
\(\dfrac{b}{2c}=\dfrac{1}{2}\)=>2b=2c =>b=c
\(\dfrac{c}{2d}=\dfrac{1}{2}\)=>2c=2d =>c=d
\(\dfrac{d}{2a}=\dfrac{1}{2}\)=>2d=2a =>d=a
=>a=b=c=d.
*\(\dfrac{2011a-2010b}{c+d}+\dfrac{2011b-2010c}{a+d}+\dfrac{2011c-2010d}{a+b}+\dfrac{2011d-2010a}{b+c}\)
=\(\dfrac{2011a-2010a}{a+a}+\dfrac{2011a-2010a}{a+a}+\dfrac{2011a-2010d}{a+a}+\dfrac{2011a-2010a}{a+a}\)
=\(\dfrac{a}{2a}+\dfrac{a}{2a}+\dfrac{a}{2a}+\dfrac{a}{2a}\)=2
\(\dfrac{2010c-2011b}{2009}=\dfrac{2011a-2009c}{2010}=\dfrac{2009b-2010a}{2011}\)
Đặt: \(\left\{{}\begin{matrix}2009=x\\2010=y\\2011=z\end{matrix}\right.\) Ta có:
\(\dfrac{cy-bz}{x}=\dfrac{az-cx}{y}=\dfrac{bx-ay}{z}\)
\(\Leftrightarrow\dfrac{cxy-bxz}{x^2}=\dfrac{ayz-cxy}{y^2}=\dfrac{bxz-ayz}{z^2}\)
Áp dụng tính chất dãy tỉ số bằng nhau:
\(\dfrac{cxy-bxz}{x^2}=\dfrac{ayz-cxy}{y^2}=\dfrac{bxz-ayz}{z^2}=\dfrac{cxy-bxz+ayz-cxy+bxz-ayz}{x^2+y^2+z^2}=0\)
\(\Leftrightarrow\left\{{}\begin{matrix}cy=bz\Leftrightarrow\dfrac{b}{y}=\dfrac{c}{z}\\az=cx\Leftrightarrow\dfrac{a}{x}=\dfrac{c}{z}\\bx=ay\Leftrightarrow\dfrac{a}{x}=\dfrac{b}{y}\end{matrix}\right.\Leftrightarrow\dfrac{a}{x}=\dfrac{b}{y}=\dfrac{c}{z}\Leftrightarrow\dfrac{a}{2009}=\dfrac{b}{2010}=\dfrac{c}{2011}\left(đpcm\right)\)