Question

tìm GTNN của B=x^2-2x+2011/x^2 với x>0

tìm GTLN của C=3(x+1)/x^3+x^2+x+1

Answer 1

1 ) \(B=\dfrac{x^2-2x+2011}{x^2}=1-\dfrac{2}{x}+\dfrac{2011}{x^2}\)

Đặt \(\dfrac{1}{x}=a\) , khi đó :

\(B=1-2a+2011a^2\)

\(=2011\left(a^2-2a.\dfrac{1}{2011}+\dfrac{1}{2011^2}\right)+\dfrac{2010}{2011}\)

\(=2011\left(a-\dfrac{1}{2011}\right)^2+\dfrac{2010}{2011}\ge\dfrac{2010}{2011}\)

Dấu " = " xảy ra \(\Leftrightarrow a=\dfrac{1}{2011}\Leftrightarrow x=2011\)

2 ) ĐKXĐ : \(x\ne-1\)\(C=\dfrac{3\left(x+1\right)}{x^3+x^2+x+1}=\dfrac{3\left(x+1\right)}{\left(x^2+1\right)\left(x+1\right)}=\dfrac{3}{x^2+1}\le\dfrac{3}{1}=3\)

Dấu " = " xảy ra \(\Leftrightarrow x=0\)