Question

Câu 1: Tìm GTLN

P=\(\dfrac{x^2}{x^4+x^2+1}\)

Answer 1

\(P=\dfrac{x^2}{x^4+x^2+1}=\dfrac{1}{x^2+1+\dfrac{1}{x^2}}\)

Áp dụng BĐT cauchy \(x^2+\dfrac{1}{x^2}\ge2\)

do đó \(P\le\dfrac{1}{3}\)

dấu = xảy ra: x=1hoặc x=-1

Answer 2

x= 0 => P = 0

xét x khác 0

\(\dfrac{1}{p}=\dfrac{x^4+x^2+1}{x^2}=x^2+\dfrac{1}{x^2}+1=\left(x^2-2+\dfrac{1}{x^2}\right)+3\)

\(\dfrac{1}{P}=\left(x-\dfrac{1}{x}\right)^2+3\ge3\) đẳng thức khi x=1 hoặc -1

\(\Rightarrow\dfrac{1}{P}\ge3\Rightarrow3P\le1\Rightarrow P\le\dfrac{1}{3}\)

Pmax =1/3 khi x=+-1