Bài 2
\(x^2+\dfrac{1}{x^2}=x^2-\dfrac{2.x.1}{x}+\dfrac{2.x.1}{x}+\dfrac{1}{x^2}=\left(x-\dfrac{1}{x}\right)^2-2\ge-2\)
Dấu ''='' xảy ra khi x = 1/x <=> x = -1 ; 1
Bài 1
\(x^2-4x+5=\left(x-2\right)^2+1\ge1\Rightarrow\dfrac{3}{\left(x-2\right)^2+1}\le3\)
Dấu ''='' xảy ra khi x = 2
\(x^2+6x+11=\left(x+3\right)^2+2\ge2\Rightarrow\dfrac{1}{\left(x+3\right)^2+2}\le\dfrac{1}{2}\)
Dấu ''='' xảy ra khi x = -3
Bài 2: \(ĐKXĐ:x\ne0\)
\(A=\dfrac{x^4+1}{x^2}=\dfrac{x^4-2x^2+1+2x^2}{x^2}=\dfrac{\left(x^2-1\right)^2+2x^2}{x^2}=\dfrac{\left(x^2-1\right)^2}{x^2}+2\ge2\)
Dấu "=" xảy ra \(\Leftrightarrow\left(x^2-1\right)^2=0\Leftrightarrow x=\pm1\)
Vậy \(MinA=2\)
