Question

Answer 1

a) Khi \(x=9\)=>\(\sqrt{x}=3\)

=>A=\(\dfrac{2.3+1}{3}=\dfrac{7}{3}\)

Vậy A=\(\dfrac{7}{3}\)khi x=9

b)\(B=\dfrac{x-3\sqrt{x}+4}{x-2\sqrt{x}}-\dfrac{1}{\sqrt{x}-2}\)

\(=\dfrac{x-3\sqrt{x}+4-\sqrt{x}}{\sqrt{x}\left(\sqrt{x}-2\right)}\)

\(=\dfrac{x-4\sqrt{x}+4}{\sqrt{x}\left(\sqrt{x}-2\right)}\)

\(=\dfrac{\left(\sqrt{x}-2\right)^2}{\sqrt{x}\left(\sqrt{x}-2\right)}=\dfrac{\sqrt{x}-2}{\sqrt{x}}\)

c)\(P=\dfrac{B}{A}=\dfrac{\sqrt{x}-2}{\sqrt{x}}.\dfrac{\sqrt{x}}{2\sqrt{x}+1}=\dfrac{\sqrt{x}-2}{2\sqrt{x}+1}\)

T có\(\left|P\right|>P\)=>\(\left[{}\begin{matrix}P>P\left(loại\right)\\-P>P\end{matrix}\right.\)

T có \(-P>P\)

\(\Rightarrow-\dfrac{\sqrt{x}-2}{2\sqrt{x}+1}>\dfrac{\sqrt{x}-2}{2\sqrt{x}+1}\)

\(\Leftrightarrow\dfrac{\sqrt{x}-2}{2\sqrt{x}+1}+\dfrac{\sqrt{x}-2}{2\sqrt{x}+1}< 0\)

\(\Leftrightarrow\dfrac{2\sqrt{x}-4}{2\sqrt{x}+1}< 0\)

Do \(x\ge0\Rightarrow2\sqrt{x}+1>0\) nên\(2\sqrt{x}-4< 0\)

\(\Rightarrow\sqrt{x}< 2\)

\(\Leftrightarrow x< 4\) \(\forall x,x\ge0\)

\(\Rightarrow0\le x< 4\)

Vậy\(\left|P\right|>P\) khi \(0\le x< 4\)