Question

tìm GTNN và GTLN của biểu thức P=\(\dfrac{x^2+1}{x^2-x+1}\)

Answer 1

\(x^2-x+1>0\)

\(P-2=\dfrac{x^2+1}{x^2-x+1}-\dfrac{2}{1}=\dfrac{x^2+1-2\left(x^2-x+1\right)}{x^2-x+1}\)

\(P-2=\dfrac{-\left(x^2-2x+1\right)}{x^2-x+1}=\dfrac{-\left(x-1\right)^2}{MSC}\le0\Rightarrow P\le2\)

\(\dfrac{2}{3}-P=\dfrac{2}{3}-\dfrac{x^2+1}{x^2-x+1}=\dfrac{2x^2-2x+2-3x^2-3}{x^2-x+1}=\dfrac{-\left(x+1\right)^2}{x^2-x+1}\le0\Rightarrow P\ge\dfrac{2}{3}\\ \)

Kết luận

GTLN P=2 khi x=-1

GTNNP =2/3 khi x=-1