Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
Cho: (2010c-2011b)/2009= (2011a-2009c)/2010= (2009b-2010a)/2011
Chứng minh rằng: a/2009=b/2010=c/2011
![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
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
\(\dfrac{a}{2b}=\dfrac{b}{2c}=\dfrac{c}{2d}=\dfrac{d}{2a}\)
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
![](https://rs.olm.vn/images/avt/0.png?1311)
Từ \(\dfrac{a}{2b}=\dfrac{b}{2c}=\dfrac{c}{2d}=\dfrac{d}{2a}\Rightarrow\dfrac{1}{2}\cdot\dfrac{a}{b}=\dfrac{1}{2}\cdot\dfrac{b}{c}=\dfrac{1}{2}\cdot\dfrac{c}{d}=\dfrac{1}{2}\cdot\dfrac{d}{a}\)
\(\Rightarrow\dfrac{a}{b}=\dfrac{b}{c}=\dfrac{c}{d}=\dfrac{d}{a}=\dfrac{a+b+c+d}{b+c+d+a}=1\)
\(\Rightarrow a=b=c=d\)
Thay \(b=a;c=a;d=a\) vào biểu thức A ta có;
\(A=\dfrac{2011a-2010a}{2a}+\)\(\dfrac{2011a-2010a}{2a}+\)\(\dfrac{2011a-2010a}{2a}+\)\(\dfrac{2011a-2010a}{2a}\)
\(A=\)\(\dfrac{a}{2a}+\)\(\dfrac{a}{2a}+\)\(\dfrac{a}{2a}+\)\(\dfrac{a}{2a}\)
\(A=\dfrac{1}{2}\cdot4=2\)
Vậy \(A=2\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Ta có :
\(B=\dfrac{2009^{2010}-2}{2009^{2011}-2}< 1\)
\(\Leftrightarrow B< \dfrac{2009^{2010}-2+2011}{2009^{2011}-2+2011}=\dfrac{2009^{2010}+2009}{2009^{2011}+2009}=\dfrac{2009\left(2009^{2009}+1\right)}{2009\left(2009^{2010}+1\right)}=\dfrac{2009^{2009}+1}{2009^{2010}+1}=A\)
\(\Leftrightarrow A>B\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(\dfrac{a}{2b}=\dfrac{b}{2c}=\dfrac{c}{2d}=\dfrac{d}{2a}=\dfrac{a+b+c+d}{2\left(a+b+c+d\right)}=\dfrac{1}{2}\)
( theo tính chất dãy tỉ số bằng nhau )
\(\Rightarrow\left\{{}\begin{matrix}a=\dfrac{1}{2}\cdot2b\\b=\dfrac{1}{2}\cdot2c\\c=\dfrac{1}{2}\cdot2d\\d=\dfrac{1}{2}\cdot2a\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}a=b\\b=c\\c=d\\d=a\end{matrix}\right.\Rightarrow a=b=c=d\)
\(\Rightarrow P=\dfrac{a}{2a}+\dfrac{a}{2a}+\dfrac{a}{2a}+\dfrac{a}{2a}=2\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Lời giải:
Áp dụng tính chất dãy tỉ số bằng nhau:
\(\frac{a}{2b}=\frac{b}{2c}=\frac{c}{2d}=\frac{d}{2a}=\frac{a+b+c+d}{2b+2c+2d+2a}=\frac{a+b+c+d}{2(a+b+c+d)}=\frac{1}{2}\)
\(\Rightarrow \frac{a}{b}=\frac{b}{c}=\frac{c}{d}=\frac{d}{a}=1\Leftrightarrow a=b=c=d\)
Do đó:
\(A=\frac{2011a-2010a}{a+a}+\frac{2011a-2010a}{a+a}+\frac{2011a-2010a}{a+a}+\frac{2011a-2010a}{a+a}\)
\(\Leftrightarrow A=\frac{1}{2}+\frac{1}{2}+\frac{1}{2}+\frac{1}{2}=2\)
Vậy \(A=2\)
Ta có: \(\dfrac{a}{2b}=\dfrac{b}{2c}=\dfrac{c}{2d}=\dfrac{d}{2a}=\dfrac{a+b+c+d}{2\left(a+b+c+d\right)}=\dfrac{1}{2}\)
\(\Rightarrow a=b;b=c;c=d;d=a\)
\(A=\dfrac{2011a-2010b}{c+d}+\dfrac{2011b-2010c}{a+d}+\dfrac{2011c-2010d}{a+b}+\dfrac{2011d-2010a}{b+c}\)
\(A=\dfrac{2011c-2010c}{c+c}+\dfrac{2011c-2010c}{c+c}+\dfrac{2011c-2010c}{c+c}+\dfrac{2011c-2010c}{c+c}\)
\(A=\dfrac{c+c+c+c}{c+c}=2\)
Vậy ....................
\(\dfrac{2010c-2011b}{2009}=\dfrac{2011a-2009c}{2010}=\dfrac{2009b-2010a}{2011}\)
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:
\(\dfrac{cxy-bxz}{x^2}=\dfrac{ayz-cxy}{y^2}=\dfrac{bxz-ayz}{z^2}=\dfrac{cxy-bxz+ayz-cxy+bxz-ayz}{x^2+y^2+z^2}=0\)
\(\Leftrightarrow\left\{{}\begin{matrix}cy=bz\Leftrightarrow\dfrac{b}{y}=\dfrac{c}{z}\\az=cx\Leftrightarrow\dfrac{a}{x}=\dfrac{c}{z}\\bx=ay\Leftrightarrow\dfrac{a}{x}=\dfrac{b}{y}\end{matrix}\right.\Leftrightarrow\dfrac{a}{x}=\dfrac{b}{y}=\dfrac{c}{z}\Leftrightarrow\dfrac{a}{2009}=\dfrac{b}{2010}=\dfrac{c}{2011}\left(đpcm\right)\)