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
