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a: Đặt a/b=c/d=k
=>a=bk; c=dk
\(\left(\dfrac{a+b}{c+d}\right)^2=\left(\dfrac{bk+b}{dk+d}\right)^2=\dfrac{b^2}{d^2}\)
\(\dfrac{a^2+b^2}{c^2+d^2}=\dfrac{b^2k^2+b^2}{d^2k^2+d^2}=\dfrac{b^2}{d^2}\)
Do đó: \(\left(\dfrac{a+b}{c+d}\right)^2=\dfrac{a^2+b^2}{c^2+d^2}\)
b: \(\dfrac{ab}{cd}=\dfrac{bk\cdot b}{dk\cdot d}=\dfrac{b^2}{d^2}\)
\(\left(\dfrac{a-b}{c-d}\right)^2=\left(\dfrac{bk-b}{dk-d}\right)^2=\dfrac{b^2}{d^2}\)
Do đó: \(\dfrac{ab}{cd}=\left(\dfrac{a-b}{c-d}\right)^2\)
\(\dfrac{2010c-2011b}{2009}=\dfrac{2011a-2009c}{2010}=\dfrac{2009b-2010a}{2011}\)
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:
\(\dfrac{cxy-bxz}{x^2}=\dfrac{ayz-cxy}{y^2}=\dfrac{bxz-ayz}{z^2}=\dfrac{cxy-bxz+ayz-cxy+bxz-ayz}{x^2+y^2+z^2}=0\)
\(\Leftrightarrow\left\{{}\begin{matrix}cy=bz\Leftrightarrow\dfrac{b}{y}=\dfrac{c}{z}\\az=cx\Leftrightarrow\dfrac{a}{x}=\dfrac{c}{z}\\bx=ay\Leftrightarrow\dfrac{a}{x}=\dfrac{b}{y}\end{matrix}\right.\Leftrightarrow\dfrac{a}{x}=\dfrac{b}{y}=\dfrac{c}{z}\Leftrightarrow\dfrac{a}{2009}=\dfrac{b}{2010}=\dfrac{c}{2011}\left(đpcm\right)\)
Ta có :
\(B=\dfrac{2009^{2010}-2}{2009^{2011}-2}< 1\)
\(\Leftrightarrow B< \dfrac{2009^{2010}-2+2011}{2009^{2011}-2+2011}=\dfrac{2009^{2010}+2009}{2009^{2011}+2009}=\dfrac{2009\left(2009^{2009}+1\right)}{2009\left(2009^{2010}+1\right)}=\dfrac{2009^{2009}+1}{2009^{2010}+1}=A\)
\(\Leftrightarrow A>B\)
a) Ta có:
\(-\dfrac{24}{35}< -\dfrac{24}{30}< -\dfrac{19}{30}\)
\(\Rightarrow x< y\)
b) Ta có:
\(A=\dfrac{2006}{2007}-\dfrac{2007}{2008}+\dfrac{2008}{2009}-\dfrac{2009}{2010}\)
\(A=\left(1-\dfrac{1}{2007}\right)-\left(1-\dfrac{1}{2008}\right)+\left(1-\dfrac{1}{2009}\right)-\left(1-\dfrac{1}{2010}\right)\)
\(A=1-\dfrac{1}{2007}-1+\dfrac{1}{2008}+1-\dfrac{1}{2009}-1+\dfrac{1}{2010}\)
\(A=-\dfrac{1}{2007}+\dfrac{1}{2008}-\dfrac{1}{2009}+\dfrac{1}{2010}\)
Ta lại có:
\(B=-\dfrac{1}{2006.2007}-\dfrac{1}{2008.2009}\)
\(B=-\dfrac{1}{2006}+\dfrac{1}{2007}-\dfrac{1}{2008}+\dfrac{1}{2009}\)
=> Dễ dàng thấy A > B
\(A=\dfrac{2006}{2007}-\dfrac{2007}{2008}+\dfrac{2008}{2009}-\dfrac{2009}{2010}\)
\(A=\left(1-\dfrac{1}{2007}\right)-\left(1-\dfrac{1}{2008}\right)+\left(1-\dfrac{1}{2009}\right)-\left(1-\dfrac{1}{2010}\right)\)
\(A=1-\dfrac{1}{2007}-1+\dfrac{1}{2008}+1-\dfrac{1}{2009}-1+\dfrac{1}{2010}\)
\(A=\left(1-1\right)+\left(1-1\right)-\dfrac{1}{2007}+\dfrac{1}{2008}-\dfrac{1}{2009}+\dfrac{1}{2010}\)
\(A=\dfrac{1}{2007}+\dfrac{1}{2008}-\dfrac{1}{2009}+\dfrac{1}{2010}\)
\(B=-\dfrac{1}{2006.2007}-\dfrac{1}{2008.2009}\)
\(B=-\left(\dfrac{1}{2006}-\dfrac{1}{2007}\right)-\left(\dfrac{1}{2008}-\dfrac{1}{2009}\right)\)
\(B=-\dfrac{1}{2006}+\dfrac{1}{2007}-\dfrac{1}{2008}+\dfrac{1}{2009}\)
\(B=\dfrac{1}{2007}+\dfrac{1}{2009}-\dfrac{1}{2006}+\dfrac{1}{2008}\)
Dễ dàng thấy \(A>B\)
\(\Leftrightarrow\left(a+2009\right)\left(b-2010\right)=\left(a-2009\right)\left(b+2010\right)\)
=>ab-2010a+2009b-2009x2010=ab+2010a-2009b-2009x2010
=>-4020a=-4018b
=>a/2009=b/2010