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\(\frac{b-2011}{c-2010}:\frac{2011-b}{2010-c}=\frac{b-2011}{c-2010}\cdot\frac{-\left(c-2010\right)}{-\left(b-2011\right)}=1\)
\(\frac{a-2009}{b-2011}=\frac{2010-c}{2009-a}=\frac{-\left(c-2010\right)}{-\left(a-2009\right)}=\frac{c-2010}{a-2009}=1\Rightarrow a-2009=c-2010=b-2011\)
\(\Rightarrow a=c-1=b-2\Rightarrow c=b-1\Rightarrow\frac{b}{c}=\frac{b}{b-1}\)=.=' ko chắc lăm
=> (a+b+c)(1/a+b +a/b+c +1/c+a)=2010 . 1/2010
=>(a+b+c) /(a+b) +(a+b+c)/(b+c) + (a+b+c)/(a+c)=1
=> 1 +c/a+b + 1 +a/b+c + 1 +b/a+c=1
=>a/b+c +b/a+c +c/a+b =-2
\(\frac{a}{b}=\frac{b}{c}=\frac{c}{2010}=\frac{2010}{a}=\frac{a+b+c+2010}{b+c+2010+a}=1\)
=>c=2010.1=2010, a=2010:1=2010, b=c=2010
Trước tiên ta có: \(\sqrt[2009]{19^{2009}+5^{2009}}>\sqrt[2009]{19^{2009}}=19\)
và \(\sqrt[2009]{19^{2009}+5^{2009}}>\sqrt[2009]{5^{2009}}=5\)
Ta có: \(\sqrt[2009]{A}=\left(19^{2009}+5^{2009}\right)\sqrt[2009]{19^{2009}+5^{2009}}\)
\(\sqrt[2009]{B}=19^{2010}+5^{2010}\)
\(\Rightarrow\sqrt[2009]{A}-\sqrt[2009]{B}=\left(19^{2009}+5^{2009}\right)\sqrt[2009]{19^{2009}+5^{2009}}-\left(19^{2010}+5^{2010}\right)\)
\(=\left(19^{2009}.\sqrt[2009]{19^{2009}+5^{2009}}-19^{2010}\right)+\left(5^{2009}.\sqrt[2009]{19^{2009}+5^{2009}}-5^{2010}\right)\)
\(=19^{2009}\left(\sqrt[2009]{19^{2009}+5^{2009}}-19\right)+5^{2009}\left(\sqrt[2009]{19^{2009}+5^{2009}}-5\right)\)
\(>19^{2009}.\left(19-19\right)+5^{2009}.\left(5-5\right)=0\)
\(\Rightarrow\sqrt[2009]{A}>\sqrt[2009]{B}\)
\(\Rightarrow A>B\)
A= (\(\left(\frac{19^{2010}}{19}+\frac{5^{2010}}{5}\right)^{2010}\)=\(\frac{\left(5.19^{2010}+19.5^{2010}\right)^{2010}}{19^{2010}.5^{2010}}\)= A(1)/A(2)
B = \(\frac{\left(19^{2010}+5^{2010}\right)^{2010}}{19^{2010}+5^{2010}}\)= B(1)/B(2)
Ta thấy A(1) >B(1), A(2)<B(2) => A>B
ủa bạn duchinhle tại sao 19^2010.5^2010 lại lớn hơn 19^2020+5^2010