Question

Giải dùm mình với ạ chi tiết nhớ cảm ơnn

 

Answer 1

a.

Khi \(x=4\Rightarrow A=\dfrac{1}{\sqrt{4}}+\dfrac{\sqrt{4}}{\sqrt{4}+1}=\dfrac{1}{2}+\dfrac{2}{3}=\dfrac{7}{6}\)

b.

\(B=\dfrac{1}{3}\Rightarrow\dfrac{\sqrt{x}}{x+\sqrt{x}}=\dfrac{1}{3}\)

\(\Rightarrow3\sqrt{x}=x+\sqrt{x}\)

\(\Rightarrow x-2\sqrt{x}=0\)

\(\Rightarrow\sqrt{x}\left(\sqrt{x}-2\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}\sqrt{x}=0\\\sqrt{x}=2\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}x=0\left(ktm\right)\\x=4\left(tm\right)\end{matrix}\right.\)

c.

\(P=A:B=\left(\dfrac{1}{\sqrt{x}}+\dfrac{\sqrt{x}}{\sqrt{x}+1}\right):\left(\dfrac{\sqrt{x}}{x+\sqrt{x}}\right)\)

\(=\left(\dfrac{\sqrt{x}+1}{\sqrt{x}\left(\sqrt{x}+1\right)}+\dfrac{x}{\sqrt{x}\left(\sqrt{x}+1\right)}\right):\left(\dfrac{\sqrt{x}}{\sqrt{x}\left(\sqrt{x}+1\right)}\right)\)

\(=\dfrac{\left(x+\sqrt{x}+1\right)}{\sqrt{x}\left(\sqrt{x}+1\right)}.\dfrac{\sqrt{x}\left(\sqrt{x}+1\right)}{\sqrt{x}}=\dfrac{x+\sqrt{x}+1}{\sqrt{x}}\)

\(P>3\Rightarrow\dfrac{x+\sqrt{x}+1}{\sqrt{x}}>3\)

\(\Leftrightarrow x+\sqrt{x}+1>3\sqrt{x}\) (do \(\sqrt{x}>0\))

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

\(\Leftrightarrow\left(\sqrt{x}-1\right)^2>0\)

\(\Leftrightarrow\sqrt{x}-1\ne0\)

\(\Rightarrow x\ne1\)

Kết hợp ĐKXĐ ta được: \(\left\{{}\begin{matrix}x>0\\x\ne1\end{matrix}\right.\)