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a) \(ĐKXĐ:x>0\)
\(Y=\frac{x^2+\sqrt{x}}{x-\sqrt{x}+1}-1-\frac{2x+\sqrt{x}}{\sqrt{x}}\)
\(\Leftrightarrow Y=\frac{\sqrt{x}\left(x\sqrt{x}+1\right)}{\left(x-\sqrt{x}+1\right)}-1-2\sqrt{x}-1\)
\(\Leftrightarrow Y=\frac{\sqrt{x}\left(\sqrt{x}+1\right)\left(x-\sqrt{x}+1\right)}{\left(x-\sqrt{x}+1\right)}-2\sqrt{x}-2\)
\(\Leftrightarrow Y=x+\sqrt{x}-2\sqrt{x}-2\)
\(\Leftrightarrow Y=x-\sqrt{x}-2\)
b) Ta có \(Y=x-\sqrt{x}-2=\left(\sqrt{x}-\frac{1}{2}\right)^2-\frac{9}{4}\ge-\frac{9}{4}\)
Dấu "=" xảy ra \(\Leftrightarrow\sqrt{x}-\frac{1}{2}=0\)
\(\Leftrightarrow x=\frac{1}{4}\)
Vậy \(Min_Y=-\frac{9}{4}\Leftrightarrow x=\frac{1}{4}\)
c) Để \(Y-\left|Y\right|=0\)
\(\Leftrightarrow Y=\left|Y\right|\)
\(\Leftrightarrow Y\ge0\)
\(\Leftrightarrow x-\sqrt{x}-2\ge0\)
\(\Leftrightarrow\left(\sqrt{x}-2\right)\left(\sqrt{x}+1\right)\ge0\)
\(\Leftrightarrow\sqrt{x}-2\ge0\) (Vì \(\sqrt{x}+1\ge0\))
\(\Leftrightarrow\sqrt{x}\ge2\)
\(\Leftrightarrow x\ge4\) (ĐPCM)
ĐKXĐ: x>0
a) \(Y=\frac{x^2+\sqrt{x}}{x-\sqrt{x}+1}+1-\frac{2x+\sqrt{x}}{\sqrt{x}}=\frac{\sqrt{x}\left(x\sqrt{x}+1\right)}{x-\sqrt{x}+1}+1-\frac{\sqrt{x}\left(2\sqrt{x}+1\right)}{\sqrt{x}}=\frac{\sqrt{x}\left(\sqrt{x}+1\right)\left(x-\sqrt{x}+1\right)}{x-\sqrt{x}+1}+1-2\sqrt{x}-1=\sqrt{x}\left(\sqrt{x}+1\right)+1-2\sqrt{x}-1=x+\sqrt{x}-2\sqrt{x}=x-\sqrt{x}\)
Ta có \(Y=x-\sqrt{x}=x-2\sqrt{x}.\frac{1}{2}+\frac{1}{4}-\frac{1}{4}=\left(\sqrt{x}-\frac{1}{2}\right)^2-\frac{1}{4}\ge-\frac{1}{4}\)
Dấu '=' xảy ra khi \(\sqrt{x}-\frac{1}{2}=0\Leftrightarrow x=\frac{1}{4}\)
Vậy GTNN của Y là \(-\frac{1}{4}\)
b) Ta có x>1\(\Leftrightarrow x>\sqrt{x}\Leftrightarrow x-\sqrt{x}>0\)
Ta lại có \(Y-\left|Y\right|=x-\sqrt{x}-\left|x-\sqrt{x}\right|=x-\sqrt{x}-\left(x-\sqrt{x}\right)=0\)
Vậy khi x>1 thì \(Y-\left|Y\right|=0\)
\(A=\left(\dfrac{1}{\sqrt{x}+\sqrt{y}}+\dfrac{1}{\sqrt{y}-\sqrt{x}}\right):\dfrac{2\sqrt{xy}}{x-y}\)
\(=\dfrac{\sqrt{x}-\sqrt{y}-\sqrt{x}-\sqrt{y}}{x-y}:\dfrac{2\sqrt{xy}}{x-y}=\dfrac{-2\sqrt{y}}{2\sqrt{xy}}=\dfrac{-1}{\sqrt{x}}=\dfrac{-\sqrt{x}}{x}\)
b, Ta có \(A=\dfrac{-1}{\sqrt{x}}=1\Leftrightarrow\sqrt{x}=-1\left(voli\right)\)
Vậy pt vô nghiệm
\(P=\left(\frac{\left(\sqrt{x}-\sqrt{y}\right)\left(1-\sqrt{xy}\right)+\left(\sqrt{x}+\sqrt{y}\right)\left(1+\sqrt{xy}\right)}{1-xy}\right):\left(\frac{x+y+2xy+1-xy}{1-xy}\right)\)
\(=\left(\frac{2\sqrt{x}+2y\sqrt{x}}{1-xy}\right):\left(\frac{\left(x+1\right)\left(y+1\right)}{1-xy}\right)\)
\(=\frac{2\sqrt{x}\left(y+1\right)}{\left(1-xy\right)}.\frac{\left(1-xy\right)}{\left(x+1\right)\left(y+1\right)}=\frac{2\sqrt{x}}{x+1}\)
\(x=\frac{2}{2+\sqrt{3}}=\frac{2\left(2-\sqrt{3}\right)}{4-3}=4-2\sqrt{3}=\left(\sqrt{3}-1\right)^2\Rightarrow\sqrt{x}=\sqrt{3}-1\)
\(\Rightarrow P=\frac{2\left(\sqrt{3}-1\right)}{5-2\sqrt{3}}=\frac{2+6\sqrt{3}}{13}\)
Ta có \(1-P=1-\frac{2\sqrt{x}}{x+1}=\frac{x-2\sqrt{x}+1}{x+1}=\frac{\left(\sqrt{x}-1\right)^2}{x+1}\ge0\) \(\forall x\ge0\)
\(\Rightarrow1-P\ge0\Rightarrow P\le1\)