ĐK: \(x\ge0;x\ne-1\)
\(a,M=\dfrac{2}{3}\left[\dfrac{1}{1+\dfrac{\left(2\sqrt{x}+1\right)^2}{3}}+\dfrac{1}{1+\dfrac{\left(2\sqrt{x}-1\right)^2}{3}}\right]\cdot\dfrac{2020}{x+1}\\ M=\dfrac{2}{3}\left[\dfrac{1}{\dfrac{3+4x+4\sqrt{x}+1}{3}}+\dfrac{1}{\dfrac{3+4x-4\sqrt{x}+1}{3}}\right]\cdot\dfrac{2020}{x+1}\\ M=\dfrac{2}{3}\left(\dfrac{3}{4x+4\sqrt{x}+4}+\dfrac{2}{4x-4\sqrt{x}+4}\right)\cdot\dfrac{2020}{x+1}\\ M=\left[\dfrac{2}{4\left(x+\sqrt{x}+1\right)}+\dfrac{2}{4\left(x-\sqrt{x}+1\right)}\right]\cdot\dfrac{2020}{x+1}\)
\(M=\dfrac{x-\sqrt{x}+1+x+\sqrt{x}+1}{2\left(x+\sqrt{x}+1\right)\left(x-\sqrt{x}+1\right)}\cdot\dfrac{2020}{x+1}\\ M=\dfrac{2\left(x+1\right)}{2\left(x+\sqrt{x}+1\right)\left(x-\sqrt{x}+1\right)}\cdot\dfrac{2020}{x+1}\\ M=\dfrac{2020}{\left(x+1\right)^2-\left(\sqrt{x}\right)^2}=\dfrac{2020}{x^2+2x+1-x}=\dfrac{2020}{x^2+x+1}\)
\(b,M=\dfrac{2020}{x^2+x+1}\)
\(M=x^2+x+1=\left(x+\dfrac{1}{2}\right)^2+\dfrac{3}{4}\ge\dfrac{3}{4}\Leftrightarrow\dfrac{2020}{x^2+x+1}\le\dfrac{2020}{\dfrac{3}{4}}=\dfrac{8080}{3}\)
Dấu \("="\Leftrightarrow x=-\dfrac{1}{2}\left(ktm\right)\)
Vậy M ko có GTLN
