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mik nghĩ đề sai lẽ ra phải là P=\(\dfrac{2010+2011\sqrt{1-x^2}+2012}{\sqrt{1-x^2}}\)(\(-1\le x\le1\))
P=\(\dfrac{2010}{\sqrt{1-x^2}}+2011+\dfrac{2012}{\sqrt{1-x^2}}=\dfrac{2010}{\sqrt{\left(1-x\right)\left(1+x\right)}}+\dfrac{2012}{\sqrt{\left(1-x\right).\left(1+x\right)}}+2011\)
áp dụng BDT CÔ SI \(\sqrt{\left(1-x\right)\left(1+x\right)}\le\dfrac{1-x+1+x}{2}=1\)
=>\(\dfrac{2010}{\sqrt{\left(1-x\right)\left(1+x\right)}}\ge2010\left(1\right)\)
tương tự \(\dfrac{2012}{\sqrt{\left(1-x\right)\left(1+x\right)}}\ge2012\left(2\right)\)
cộng vế (1)(2)=>\(\dfrac{2010}{\sqrt{\left(1-x\right)\left(1+x\right)}}+\dfrac{2012.}{\sqrt{\left(1-x\right)\left(1+x\right)}}\ge2012+2010=4022\)
=>\(\dfrac{2010}{\sqrt{\left(1-x\right)\left(1+x\right)}}+\dfrac{2012}{\sqrt{\left(1+x\right)\left(1-x\right)}}+2011\ge4022+2011=6033\)
dấu = xảy ra khi và chỉ khi x=0
vậy min P=6033
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\(N=\left(2010^{2010}+2011^{2010}\right)^{2011}=\left(2010^{2010}+2011^{2010}\right)^{2010}.\left(2010^{2010}+2011^{2010}\right)\)
\(>\left(2010^{2010}+2011^{2010}\right)^{2010}.2011^{2010}=\left[\left(2010^{2010}+2011^{2010}\right)2011\right]^{2010}\)
\(>\left(2010^{2010}.2010+2011^{2010}.2011\right)^{2010}=\left(2010^{2011}+2011^{2011}\right)^{2010}=M\)
Vậy M < N,
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Đặt 2011=t
\(\Rightarrow T=\sqrt{1+\left(t-1\right)^2+\frac{\left(t-1\right)^2}{t^2}}+\frac{t-1}{t}\)
\(=\sqrt{\frac{t^2+t^2\left(t-1\right)^2+\left(t-1\right)^2}{t^2}}+\frac{t-1}{t}\)
\(=\frac{\sqrt{t^2+t^4-2t^3+t^2+t^2-2t+1}+t-1}{t}\)
\(=\frac{\sqrt{t^4+t^2+1+2t^2-2t^3-2t}+t-1}{t}\)
\(=\frac{\sqrt{\left(t^2-t+1\right)^2}+t-1}{t}\)
\(=\frac{t^2-t+1+t-1}{t}=t=2011\)
mà \(2011\in Z\)
nên T là một số nguyên.