Ta có:\(\frac{a}{b}=\frac{c}{d}\Rightarrow\frac{a}{c}=\frac{b}{d}=\frac{a-b}{c-d}\)

\(\Rightarrow\left(\frac{a}{c}\right)^{2008}=\left(\frac{b}{d}\right)^{2008}=\left(\frac{a-b}{c-d}\right)^{2008}=\frac{a^{2008}}{c^{2008}}=\frac{b^{2008}}{d^{2008}}=\frac{a^{2008}+b^{2008}}{c^{2008}+d^{2008}}\)

\(\Rightarrow\left(\frac{a-b}{c-d}\right)^{2008}=\frac{a^{2008}+b^{2008}}{c^{2008}+d^{2008}}\left(đpcm\right)\)

1003/1004

\(\frac{2008}{2009}-\frac{2009}{2008}+\frac{1}{2009}+\frac{2007}{2008}=\frac{1003}{1004}\)

ai k mình mình k lại,ok

\(\frac{1003}{1004}\)

\(a-b+c+d=\frac{2008}{2009}-\frac{2009}{2008}+\frac{1}{2009}+\frac{2007}{2008}\)

\(=\left(\frac{2008}{2009}+\frac{1}{2009}\right)-\left(\frac{2009}{2008}-\frac{2007}{2008}\right)\)

\(=1-\frac{2}{2008}\)

\(=\frac{1003}{1004}\)

Đpcm

⇔ \(\dfrac{a+b+c-a}{a}+\dfrac{a+b+c-b}{b}+\dfrac{a+b+c-c}{c}\) ≥ 6

⇔ \(\dfrac{b+c}{a}+\dfrac{a+b}{c}+\dfrac{a+c}{b}\ge6\)

⇔ \(\dfrac{b}{a}+\dfrac{c}{a}+\dfrac{c}{b}+\dfrac{a}{b}+\dfrac{a}{c}+\dfrac{b}{c}\ge6\) (1)

Bất đẳng thức Cosi => (1)

Dấu bằng xảy ra khi a = b = c = \(\dfrac{2008}{3}\)

a-b+c+d=\(\frac{2008}{2009}-\frac{2009}{2008}+\frac{1}{2009}+\frac{2007}{2008}=\left(\frac{2008}{2009}+\frac{1}{2009}\right)-\left(\frac{2009}{2008}-\frac{2007}{2008}\right)=1-\frac{2}{2008}=\frac{2006}{2008}=\frac{1003}{1004}\)

\(a-b+c+d=\frac{2008}{2009}-\frac{2009}{2008}+\frac{1}{2009}+\frac{2007}{2008}\)

\(=\left(\frac{2008}{2009}+\frac{1}{2009}\right)+\left(\frac{2007}{2008}-\frac{2009}{2008}\right)=\frac{2009}{2009}+\frac{-2}{2008}\)

\(=1+\frac{-1}{1004}=\frac{1004}{1004}+\frac{-1}{1004}=\frac{1003}{1004}\)

Câu 18: B

Câu 19: C