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a) đk: \(x\ge0;x\ne\left\{\frac{1}{4};1\right\}\)
\(P=\left(\frac{2x\sqrt{x}+x-\sqrt{x}}{x\sqrt{x}-1}-\frac{x+\sqrt{x}}{x-1}\right)\cdot\frac{x-1}{2x+\sqrt{x}-1}+\frac{\sqrt{x}}{2\sqrt{x}-1}\)
\(P=\left[\frac{\left(2x+\sqrt{x}-1\right)\sqrt{x}}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}-\frac{\left(\sqrt{x}+1\right)\sqrt{x}}{x-1}\right]\cdot\frac{x-1}{2x+\sqrt{x}-1}+\frac{\sqrt{x}}{2\sqrt{x}-1}\)
\(P=\frac{\left(x-1\right)\sqrt{x}}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}-\frac{\left(\sqrt{x}+1\right)\sqrt{x}}{2x+\sqrt{x}-1}+\frac{\sqrt{x}}{2\sqrt{x}-1}\)
\(P=\frac{x+\sqrt{x}}{x+\sqrt{x}+1}-\frac{\sqrt{x}}{2\sqrt{x}-1}+\frac{\sqrt{x}}{2\sqrt{x}-1}\)
\(P=\frac{x+\sqrt{x}}{x+\sqrt{x}+1}\)
b) Ta có:
\(P=\frac{x+\sqrt{x}}{x+\sqrt{x}+1}=\frac{\left(x+\sqrt{x}+1\right)-1}{x+\sqrt{x}+1}=1-\frac{1}{x+\sqrt{x}+1}\)
Mà \(x+\sqrt{x}\ge0\left(\forall x\right)\)
\(\Leftrightarrow x+\sqrt{x}+1\ge1\left(\forall x\right)\)
\(\Leftrightarrow\frac{1}{x+\sqrt{x}+1}\le1\left(\forall x\right)\)
\(\Leftrightarrow P=1-\frac{1}{x+\sqrt{x}+1}\ge0\left(\forall x\right)\)
Dấu "=" xảy ra khi: \(x+\sqrt{x}=0\Leftrightarrow x=0\)
Vậy Min(P) = 0 khi x = 0
a.\(DK:x\ge0\)
\(A=\frac{x-2\sqrt{x}+1}{x+1}.\frac{\left(x+1\right)\left(\sqrt{x}+1\right)}{x-2\sqrt{x}+1}=\sqrt{x}+1\)
b.Dat \(P=\frac{1}{A}\left(x+3\right)=\frac{x+3}{\sqrt{x}+1}\left(P>0\right)\)
\(\Rightarrow P\sqrt{x}+P=x+3\)
\(\Leftrightarrow x-P\sqrt{x}+3-P=0\)
Dat \(t=\sqrt{x}\left(t\ge0\right)\)
Ta co:
\(\Delta\ge0\)
\(\Leftrightarrow P^2-4\left(3-P\right)\ge0\)
\(\Leftrightarrow P^2+4P-12\ge0\)
\(\Leftrightarrow\left(P-2\right)\left(P+6\right)\ge0\)
TH1:
\(\hept{\begin{cases}P-2\ge0\\P+6\ge0\end{cases}\Leftrightarrow P\ge2}\)
TH2:
\(\hept{\begin{cases}P-2\le0\\P+6\le0\end{cases}\Leftrightarrow P\le2\left(P>0\right)}\)
Vi la de bai tim min nen lay TH1 thoi
Dau '=' xay ra khi \(x=\frac{P}{2}=1\)
Vay \(P_{min}=2\)khi \(x=1\)