ĐK : \(x>0\)
Câu a : \(P=\left(\dfrac{1}{\sqrt{x}}+\dfrac{\sqrt{x}}{\sqrt{x}+1}\right):\dfrac{\sqrt{x}}{x+\sqrt{x}}\)
\(=\dfrac{\sqrt{x}+1+x}{\sqrt{x}\left(\sqrt{x}+1\right)}:\dfrac{\sqrt{x}}{\sqrt{x}\left(\sqrt{x}+1\right)}\)
\(=\dfrac{x+\sqrt{x}+1}{\sqrt{x}\left(\sqrt{x}+1\right)}\times\dfrac{\sqrt{x}\left(\sqrt{x}+1\right)}{\sqrt{x}}\)
\(=\dfrac{x+\sqrt{x}+1}{\sqrt{x}}\)
Câu b : Thay \(x=4\) vào biểu thức P ta có :
\(P=\dfrac{4+\sqrt{4}+1}{\sqrt{4}}=\dfrac{4+2+1}{2}=\dfrac{7}{2}=3,5\)
Câu c : \(P=\dfrac{13}{3}\Leftrightarrow\dfrac{x+\sqrt{x}+1}{\sqrt{x}}=\dfrac{13}{3}\)
\(\Leftrightarrow3x+3\sqrt{x}+3=13\sqrt{x}\)
\(\Leftrightarrow3x-10\sqrt{x}+3=0\)
\(\Leftrightarrow3x-\sqrt{x}-9\sqrt{x}+3=0\)
\(\Leftrightarrow\sqrt{x}\left(3\sqrt{x}-1\right)-3\left(3\sqrt{x}-1\right)=0\)
\(\Leftrightarrow\left(3\sqrt{x}-1\right)\left(\sqrt{x}-3\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}3\sqrt{x}-1=0\\\sqrt{x}-3=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1}{9}\\x=9\end{matrix}\right.\)
Chúc bạn học tốt !
