đặt a^2010+2009=b
\(\Rightarrow\frac{b+1}{\sqrt{b}}\)
ta có : b+1\(\ge\)2\(\sqrt{b}\) ( cô - si)
\(\frac{b+1}{\sqrt{b}}\ge2\)
dấu = xảy ra \(\Leftrightarrow b=1\)
\(\Rightarrowđpcm\)
Chứng minh\(\frac{2010}{\sqrt{2009}}+\frac{2009}{\sqrt{2010}}>\sqrt{2009}+\sqrt{2010}\)
\(\frac{a^{2010}+2010}{\sqrt{a^{2010}+2009}}\ge2\)
tính b=\(1^2-2^2+3^2-...+2008^2-2009^2\)
a=\(\frac{1}{2\sqrt{1}+1\sqrt{2}}+\frac{1}{3\sqrt{2}+2\sqrt{3}}+....+\frac{1}{2010\sqrt{2009}+2009\sqrt{2010}}\)
tính U =(\(\left(\frac{1}{3\sqrt{2}+2\sqrt{3}}+\frac{1}{4\sqrt{3}+3\sqrt{4}}+...+\frac{1}{2010\sqrt{2009}+2009\sqrt{2010}}\right)\)
giải pt:\(\frac{\sqrt{x-2009}-1}{x-2009}+\frac{\sqrt{y-2010}-1}{y-2010}+\frac{\sqrt{z-2011}-1}{z-2011}=\frac{3}{4}\)
Giải phương trình
\(\frac{\sqrt{x-2009}}{x-2009}+\frac{\sqrt{y-2010}}{y-2010}+\frac{\sqrt{z-2011}}{z-2011}=\frac{3}{4}\)
Giải phương trình: \(\frac{\sqrt{x-2009}-1}{x-2009}+\frac{\sqrt{y-2010}-1}{y-2010}+\frac{\sqrt{z-2011}-1}{z-2011}=\frac{3}{4}\)
Giải phương trình :
\(\frac{\sqrt{x-2009}-1}{x-2009}+\frac{\sqrt{y-2010}-1}{y-2010}+\frac{\sqrt{z-2011}-1}{z-2011}=\frac{3}{4}\)
Giải phương trình: \(\frac{\sqrt{x-2009}-1}{x-2009}+\frac{\sqrt{y-2010}-1}{y-2010}+\frac{\sqrt{z-2011}-1}{z-2011}=\frac{3}{4}\)
