Question

Answer 1

a, C xác định khi \(\left\{{}\begin{matrix}x\ge0\\\sqrt{x}-1\ne0\\x+\sqrt{x}\ne0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x>0\\x\ne1\end{matrix}\right.\)

b, \(C=\left(\dfrac{x-\sqrt{x}}{\sqrt{x}-1}+1\right):\left(\dfrac{x+\sqrt{x}}{\sqrt{x}+1}\right)\)

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

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

\(=\dfrac{\sqrt{x}+1}{\sqrt{x}}\)

Answer 2

\(a.x>1\)

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

c.Thay \(x=\dfrac{4}{9}\) vào biểu thức C đã rút gọn ta có:

\(\left(\dfrac{4}{9}+\sqrt{\dfrac{4}{9}}\right):\dfrac{4}{9}\\ =\dfrac{5}{2}\)

 

Answer 3

a. Để \(C\) có nghĩa thì \(\left\{{}\begin{matrix}\sqrt{x}\ge0\\\sqrt{x}-1>0\\\sqrt{x}+1>0\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x\ge0\\x>1\end{matrix}\right.\) \(\Leftrightarrow x>1\)

b. \(C=\left(\dfrac{x-\sqrt{x}}{\sqrt{x}-1}+1\right):\left(\dfrac{x+\sqrt{x}}{\sqrt{x}+1}\right)\)

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

\(=\dfrac{\sqrt{x}+1}{\sqrt{x}}\)

c. Thay \(x=\dfrac{4}{9}\) vào \(C\), ta được

\(C=\dfrac{\sqrt{\dfrac{4}{9}}+1}{\sqrt{\dfrac{4}{9}}}=\dfrac{\dfrac{2}{3}+1}{\dfrac{2}{3}}=\dfrac{\dfrac{5}{3}}{\dfrac{2}{3}}=\dfrac{5}{2}\)

d. Để \(C=5\) thì \(\dfrac{\sqrt{x}+1}{\sqrt{x}}=5\Leftrightarrow5\sqrt{x}=\sqrt{x}+1\Leftrightarrow4\sqrt{x}=1\Leftrightarrow\sqrt{x}=\dfrac{1}{4}\Leftrightarrow x=\dfrac{1}{16}\)

Vậy x thỏa ycbt là \(x=\dfrac{1}{16}\)

 

Answer 4

a: ĐKXĐ: \(\left\{{}\begin{matrix}x>0\\x\ne1\end{matrix}\right.\)

b: Ta có: \(C=\left(\dfrac{x-\sqrt{x}}{\sqrt{x}-1}+1\right):\dfrac{x+\sqrt{x}}{\sqrt{x}+1}\)

\(=\dfrac{\sqrt{x}+1}{\sqrt{x}}\)

c: Thay \(x=\dfrac{4}{9}\) vào C, ta được:

\(C=\left(\dfrac{2}{3}+1\right):\dfrac{2}{3}=\dfrac{5}{3}\cdot\dfrac{3}{2}=\dfrac{5}{2}\)

d: Để C=5 thì \(\sqrt{x}+1=5\sqrt{x}\)

\(\Leftrightarrow\sqrt{x}=\dfrac{1}{4}\)

hay \(x=\dfrac{1}{16}\)