19) a) Không ghi lại biểu thức đề cho
\(P=\left[\dfrac{2\sqrt{x}\left(\sqrt{x}-3\right)+\sqrt{x}\left(3+\sqrt{x}\right)-3x-3}{\left(3+\sqrt{x}\right)\left(\sqrt{x}-3\right)}\right]:\dfrac{2\sqrt{x}-2-\sqrt{x}+3}{\sqrt{x}-3}\)
\(P=\left[\dfrac{2x-6\sqrt{x}+3\sqrt{x}+x-3x-3}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}\right]:\dfrac{\sqrt{x}+1}{\sqrt{x}-3}\)
\(P=\left[\dfrac{-3\sqrt{x}-3}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}\right].+\dfrac{\sqrt{x}-3}{\sqrt{x}+1}\)
\(P=\dfrac{-3\left(\sqrt{x}+1\right)}{\left(\sqrt{x}+3\right)\left(\sqrt{x}+1\right)}\)
\(P=\dfrac{-3}{\sqrt{x}+3}\)
b) ĐK: \(x\ge0\)
\(P< \dfrac{-1}{3}\)
\(\Leftrightarrow\dfrac{-3}{\sqrt{x}+3}< \dfrac{-1}{3}\)
\(\Leftrightarrow-9< -\sqrt{x}-3\)
\(\Leftrightarrow\sqrt{x}< 6\)
\(\Leftrightarrow x< 36\)
Vậy: \(0\le x< 36\) thì \(P< \dfrac{-1}{3}\)
c) \(\sqrt{x}\ge0\Rightarrow\sqrt{x}+3\ge3\)
\(\Rightarrow\dfrac{1}{\sqrt{x}+3}\le\dfrac{1}{3}\)
\(\Rightarrow\dfrac{-3}{\sqrt{x}+3}\ge-1\)
Vậy: MinP=-1 khi x=0
Bạn ở thái Nguyên ?
\(\left(\dfrac{2\sqrt{x}}{3+\sqrt{x}}+\dfrac{\sqrt{x}}{\sqrt{x}-3}-\dfrac{3x+3}{x-9}\right):\left(\dfrac{2\sqrt{x}}{\sqrt{x}-3}-1\right)\)<=>\(\dfrac{2x-6\sqrt{x}+3\sqrt{x}+x-3x-3}{x-9}:\dfrac{\sqrt{x}+3}{\sqrt{x}-3}\)
<=>\(\dfrac{\left(-3\sqrt{x}-3\right)\left(\sqrt{x}-3\right)}{\left(x-9\right)\left(\sqrt{x}+3\right)}=\dfrac{-3\left(\sqrt{x-1}\right)}{\left(\sqrt{x}+3\right)^2}\)
b)\(\dfrac{-3\left(\sqrt{x}+1\right)}{\left(\sqrt{x}+3\right)^2}=\dfrac{-3\sqrt{x}-3}{\left(\sqrt{x}+3\right)^2}\)
=\(\dfrac{\dfrac{3}{8}\left(-8\sqrt{x}-8\right)}{\left(\sqrt{x}+3\right)^2}\)=\(\dfrac{\dfrac{3}{8}\left(-x+x-8x-8\right)}{\left(\sqrt{x}+3\right)^2}\)=\(\dfrac{\dfrac{3}{8}\left[\left(-x-6x-9\right)+\left(x-2x+1\right)\right]}{\left(\sqrt{x}+3\right)^2}\)
=\(\dfrac{\dfrac{3}{8}\left[-\left(\sqrt{x}+3\right)^2+\left(\sqrt{x}-1\right)^2\right]}{\left(\sqrt{x}+3\right)^2}=\dfrac{\dfrac{3}{8}\left(\sqrt{x}-1\right)^2}{\left(\sqrt{x}+3\right)^2}+\dfrac{-3}{8}\ge\dfrac{-3}{8}\)
Dấu"=" xảy ra khi \(\sqrt{x}-1=0\)=> x=1
