Áp dụng BĐT cosi:

\(A=\sqrt{\left(2x+1\right)\left(x+2\right)}+2\sqrt{x+3}-2x\\ A\le\dfrac{2x+1+x+2}{2}+\dfrac{4+x+3}{2}-2x\\ A\le\dfrac{3x+3}{2}+\dfrac{x+7}{2}-2x=\dfrac{3x+3+x+7-4x}{2}=5\)

Dấu \("="\Leftrightarrow\left\{{}\begin{matrix}2x+1=x+2\\4=x+3\end{matrix}\right.\Leftrightarrow x=1\)

\(P=\sqrt{\left(x+2\right)\left(2x+1\right)}+2\sqrt{x+3}-2x\)

\(P\le\dfrac{1}{2}\left(x+2+2x+1\right)+\dfrac{1}{2}\left(4+x+3\right)-2x=5\)

\(P_{max}=5\) khi \(x=1\)

\(5x^2+2xy+2y^2-\left(4x^2+4xy+y^2\right)=\left(x-y\right)^2\ge0\\ \Leftrightarrow5x^2+2xy+2y^2\ge4x^2+4xy+y^2=\left(2x+y\right)^2\)

\(\Leftrightarrow P\le\dfrac{1}{2x+y}+\dfrac{1}{2y+z}+\dfrac{1}{2z+x}=\dfrac{1}{9}\left(\dfrac{9}{x+x+y}+\dfrac{9}{y+y+z}+\dfrac{9}{z+z+x}\right)\\ \Leftrightarrow P\le\dfrac{1}{9}\left(\dfrac{1}{x}+\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{y}+\dfrac{1}{y}+\dfrac{1}{z}+\dfrac{1}{z}+\dfrac{1}{z}+\dfrac{1}{x}\right)\\ \Leftrightarrow P\le\dfrac{1}{9}\left(\dfrac{3}{x}+\dfrac{3}{y}+\dfrac{3}{z}\right)=\dfrac{1}{3}\left(\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}\right)=1\)

Dấu \("="\Leftrightarrow x=y=z=1\)

\(\sqrt{5x^2+2xy+2y^2}=\sqrt{4x^2+2xy+y^2+x^2+y^2}\ge\sqrt{4x^2+2xy+y^2+2xy}=2x+y\)

\(\Rightarrow\dfrac{1}{\sqrt{5x^2+2xy+2y^2}}\le\dfrac{1}{2x+y}=\dfrac{1}{x+x+y}\le\dfrac{1}{9}\left(\dfrac{1}{x}+\dfrac{1}{x}+\dfrac{1}{y}\right)=\dfrac{1}{9}\left(\dfrac{2}{x}+\dfrac{1}{y}\right)\)

Tương tự:

\(\dfrac{1}{\sqrt{5y^2+2yz+2z^2}}\le\dfrac{1}{9}\left(\dfrac{2}{y}+\dfrac{1}{z}\right)\) ; \(\dfrac{1}{\sqrt{5z^2+2zx+2x^2}}\le\dfrac{1}{9}\left(\dfrac{2}{z}+\dfrac{1}{x}\right)\)

Cộng vế:

\(P\le\dfrac{1}{9}\left(\dfrac{3}{x}+\dfrac{3}{y}+\dfrac{3}{z}\right)=1\)

\(P_{max}=1\) khi \(x=y=z=1\)

GTNN và GTLN của cả A và B hay của A + B vậy bạn...

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

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

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

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

b. Đặt \(B=A-2x\)

\(B=\sqrt{x}-1-2x=-2\left(\sqrt{x}-\dfrac{1}{4}\right)^2-\dfrac{7}{8}\le-\dfrac{7}{8}\)

\(B_{max}=-\dfrac{7}{8}\) khi \(\sqrt{x}-\dfrac{1}{4}=0\Leftrightarrow x=\dfrac{1}{16}\)

\(F=1-\sqrt{x^2-2x+2}=1-\sqrt{\left(x-1\right)^2+1}\)(   Điều kiện: \(x\in R\))

Ta có \(\left(x-1\right)^2\ge0, \forall x \Leftrightarrow\left(x-1\right)^2+1\ge1, \forall x \Leftrightarrow\sqrt{\left(x-1\right)^2+1} \ge1, \forall x\)

\(\Leftrightarrow-\sqrt{\left(x-1\right)^2+1}\le-1, \forall x \Leftrightarrow1-\sqrt{\left(x-1\right)^2+1}\le0, \forall x\Leftrightarrow F\le0, \forall x\)

Dấu "=" xảy ra \(\Leftrightarrow\left(x-1\right)^2=0\Leftrightarrow x=1\)( thỏa điều kiện )

Vậy GTLN của F là 0 tại x = 1

dệ không