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1.
\(\sqrt{\dfrac{x-1+\sqrt{2x-3}}{x+2-\sqrt{2x+3}}}\Leftrightarrow\)\(\left\{{}\begin{matrix}x\ge\dfrac{3}{2}\\\sqrt{\dfrac{\left(\sqrt{2x-3}+1\right)^2}{\left(\sqrt{2x+3}-1\right)^2}}\end{matrix}\right.\)\(\Leftrightarrow\)
\(\Leftrightarrow\left\{{}\begin{matrix}x\ge\dfrac{3}{2}\\\dfrac{\sqrt{2x-3}+1}{\sqrt{2x+3}-1}\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}x\ge\dfrac{3}{2}\\\dfrac{\left(\sqrt{2x-3}+1\right)\left(\sqrt{2x+3}+1\right)}{2\left(x+1\right)}\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x\ge\dfrac{3}{2}\\\dfrac{\sqrt{4x^2-9}+\sqrt{2x-3}+\sqrt{2x+3}+1}{2\left(x+1\right)}\end{matrix}\right.\)
hết tối giải rồi
\(F=\dfrac{\sqrt{x}-\sqrt{y}}{xy\sqrt{xy}}:\left[\dfrac{x+y}{xy}\cdot\dfrac{1}{\left(\sqrt{x}+\sqrt{y}\right)^2}+\dfrac{2}{\sqrt{xy}\left(\sqrt{x}+\sqrt{y}\right)^2}\right]\)
\(=\dfrac{\sqrt{x}-\sqrt{y}}{xy\sqrt{xy}}:\left[\dfrac{x+y+2\sqrt{xy}}{xy\left(\sqrt{x}+\sqrt{y}\right)^2}\right]\)
\(=\dfrac{\sqrt{x}-\sqrt{y}}{xy\sqrt{xy}}\cdot xy=\dfrac{\sqrt{x}-\sqrt{y}}{\sqrt{xy}}\)
\(\left(\sqrt{x}+\sqrt{y}\right)^2-\sqrt{3}\left(\sqrt{x}+\sqrt{y}\right)-\sqrt{xy}\)
\(=x+2\sqrt{xy}+y-\sqrt{3x}-\sqrt{3y}-\sqrt{xy}\)
\(=\left(\sqrt{x}-\sqrt{y}\right)^2-\sqrt{3x}-\sqrt{3y}+3\sqrt{xy}\)
\(=\left(\sqrt{x}-\sqrt{y}\right)^2+\sqrt{3x}\left(\sqrt{3y}-1\right)-\sqrt{3y}+1-1\)
\(=\left(\sqrt{x}-\sqrt{y}\right)^2+\left(\sqrt{3x}-1\right)\left(\sqrt{3y}-1\right)-1\)\(\ge-1\forall x,y\)
Dấu "=" xảy ra <=> x=y=1/3