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22 tháng 2 2017

\(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=\frac{1}{a+b+c}\)

\(\Leftrightarrow\frac{1}{a}+\frac{1}{b}+\frac{1}{c}-\frac{1}{a+b+c}=0\)

\(\Leftrightarrow\frac{a+b}{ab}+\frac{a+b+c-c}{c\left(a+b+c\right)}=0\)

\(\Leftrightarrow\frac{a+b}{ab}+\frac{a+b}{c\left(a+b+c\right)}=0\)

\(\Leftrightarrow\left(a+b\right)\left(\frac{1}{ab}+\frac{1}{c\left(a+b+c\right)}\right)=0\)

\(\Leftrightarrow\left(a+b\right)\left(\frac{ca+cb+c^2+ab}{abc\left(a+b+c\right)}\right)=0\)

\(\Leftrightarrow\left(a+b\right)\left(b\left(a+c\right)+c\left(a+c\right)\right)=0\)

\(\Leftrightarrow\left(a+b\right)\left(b+c\right)\left(a+c\right)=0\)

\(\Rightarrow a+b=0\Rightarrow a=-b\Rightarrow a^{2009}=-b^{2009}\)

\(\frac{1}{a^{2009}}+\frac{1}{b^{2009}}+\frac{1}{c^{2009}}=\frac{1}{c^{2009}}\) (1)

\(\frac{1}{a^{2009}+b^{2009}+c^{2009}}=\frac{1}{c^{2009}}\) (2)

Từ (1) và (2) \(\Rightarrow\frac{1}{a^{2009}}+\frac{1}{b^{2009}}+\frac{1}{c^{2009}}=\frac{1}{a^{2009}+b^{2009}+c^{2009}}\) (đpcm)

\(\dfrac{a+c}{b+d}=\dfrac{a-c}{b-d}\)

\(\Leftrightarrow ab-ad+cb-cd=ab+ad-cb-cd\)

=>-2ad=-2cb

=>ad=cb

=>a/b=c/d

Đặt a/b=c/d=k

=>a=bk; c=dk

\(\dfrac{a^{2009}-c^{2009}}{b^{2009}-d^{2009}}=\dfrac{b^{2009}k^{2009}-d^{2009}k^{2009}}{b^{2009}-d^{2009}}=k^{2009}\)

\(\left(\dfrac{a}{b}\right)^{2009}=\left(\dfrac{bk}{b}\right)^{2009}=k^{2009}\)

Do đó: \(\dfrac{a^{2009}-c^{2009}}{b^{2009}-d^{2009}}=\left(\dfrac{a}{b}\right)^{2009}\)