Question

Cho M=\(\dfrac{x^2-2x}{x^3+1}+\dfrac{1}{2}\left(\dfrac{1}{1+\sqrt{x+2}}+\dfrac{1}{1-\sqrt{x+2}}\right)\)

a) Rút gọn M

b)Tìm GTNN của M

Answer 1

ĐKXĐ: \(x\ge-2;x\ne-1\)

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

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

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

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

\(M_{min}=-\dfrac{4}{3}\) khi \(x=\dfrac{1}{2}\)