1) A = 4 khi:
\(\dfrac{x-4}{\sqrt{x}-1}=4\)
\(\Leftrightarrow4\sqrt{x}-4=x-4\)
\(\Leftrightarrow4\sqrt{x}-x=-4+4\)
\(\Leftrightarrow\sqrt{x}\left(4-\sqrt{x}\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\left(tm\right)\\x=16\left(tm\right)\end{matrix}\right.\)
2) \(B=\left(\dfrac{\sqrt{x}-1}{\sqrt{x}-2}-\dfrac{\sqrt{x}+2}{\sqrt{x}+1}\right):\dfrac{3}{\sqrt{x}+1}\)
\(B=\left[\dfrac{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+1\right)}-\dfrac{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+1\right)}\right]\cdot\dfrac{\sqrt{x}+1}{3}\)
\(B=\dfrac{x-1-x+4}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+1\right)}\cdot\dfrac{\sqrt{x}+1}{3}\)
\(B=\dfrac{3\left(\sqrt{x}+1\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+1\right)3}\)
\(B=\dfrac{1}{\sqrt{x}-2}\)
1) Đk: \(x\ge0;x\ne1\)
\(A=4\Leftrightarrow\dfrac{x-4}{\sqrt{x}-1}=4\Leftrightarrow\dfrac{x-4}{\sqrt{x}-1}-4=0\)
\(\Rightarrow x-4-4\left(\sqrt{x}-1\right)=0\Leftrightarrow x-4\sqrt{x}=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=16\end{matrix}\right.\) (tm)
Vậy...
2) ĐK: \(x\ge0;x\ne4\)
\(B=\dfrac{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)-\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+1\right)}.\dfrac{\sqrt{x}+1}{3}\)
\(=\dfrac{x-1-\left(x-4\right)}{3\left(\sqrt{x}-2\right)}=\dfrac{1}{\sqrt{x}-2}\)
Vậy...
3) \(P=\dfrac{1}{AB}=\dfrac{1}{\dfrac{x-4}{\sqrt{x}-1}.\dfrac{1}{\sqrt{x}-2}}\)\(=\dfrac{\left(\sqrt{x}-1\right)\left(\sqrt{x}-2\right)}{x-4}=\dfrac{\sqrt{x}-1}{\sqrt{x}+2}\)\(=1-\dfrac{3}{\sqrt{x}+2}\)
Có \(\sqrt{x}+2\ge2\Leftrightarrow\dfrac{3}{\sqrt{x}+2}\le\dfrac{3}{2}\) \(\Leftrightarrow1-\dfrac{3}{\sqrt{x}+2}\ge1-\dfrac{3}{2}\Leftrightarrow P\ge-\dfrac{1}{2}\)
Suy ra \(P_{min}=-\dfrac{1}{2}\Leftrightarrow x=0\)
Vậy...
4) \(P-\dfrac{3}{2}=1-\dfrac{3}{\sqrt{x}+2}-\dfrac{3}{2}=-\dfrac{1}{2}-\dfrac{3}{\sqrt{x}+2}< 0\)
Suy ra \(P< \dfrac{3}{2}\)
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