ĐKXĐ: \(\frac{1}{\sqrt{x}-1}-\frac{1}{\sqrt{x}+1}+1\)
K
Khách
Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
Những câu hỏi liên quan
LM
8 tháng 11 2020
A=\(\frac{x}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}+\frac{1}{\sqrt{x}-2}+\frac{1}{\sqrt{x}+2}\)
=\(\frac{x+\sqrt{x}+2+\sqrt{x}-2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}=\frac{x+2\sqrt{x}}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\)
=\(\frac{\sqrt{x}\left(\sqrt{x}+2\right)}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}=\frac{\sqrt{x}}{\sqrt{x-2}}\)
Vậy A=\(\frac{\sqrt{x}}{\sqrt{x}-2}\)vs x\(\ge0;x\ne4\)
LM
9 tháng 11 2020
C=\(\left(\frac{1+x}{\sqrt{x}\left(\sqrt{x}+1\right)}\right)\times\frac{\sqrt{x}\left(\sqrt{x}+1\right)}{\sqrt{x}}=\frac{1+x}{\sqrt{x}}\)
Vậy C=\(\frac{1+x}{\sqrt{x}}\)vs x>0
26 tháng 7 2018
A = \(\frac{1+x}{x+\sqrt{x}}.\frac{\sqrt{x}+1}{3}\)=\(\frac{1+x}{3\sqrt{x}}\)
ĐKXĐ : x > 0
TB
25 tháng 7 2018
ĐKXĐ: \(x\ge0\)
\(\frac{1}{\sqrt{x}+1}-\frac{3}{x\sqrt{x}+1}+\frac{2}{x-\sqrt{x}+1}\)
\(=\frac{1}{\sqrt{x}+1}-\frac{3}{\left(\sqrt{x}+1\right)\left(x-\sqrt{x}+1\right)}+\frac{2}{x-\sqrt{x}+1}\)
\(=\frac{x-\sqrt{x}+1-3+2\sqrt{x}+2}{\left(\sqrt{x}+1\right)\left(x-\sqrt{x}+1\right)}\)
\(=\frac{x+\sqrt{x}}{\left(\sqrt{x}+1\right)\left(x-\sqrt{x}+1\right)}\)
\(=\frac{\sqrt{x}\left(\sqrt{x}+1\right)}{\left(\sqrt{x}+1\right)\left(x-\sqrt{x}+1\right)}\)
\(=\frac{\sqrt{x}}{x-\sqrt{x}+1}\)
15 tháng 7 2019
\(ĐKXĐ:\hept{\begin{cases}\sqrt{x}-1\ne0\\x\ge0\end{cases}\Leftrightarrow}\hept{\begin{cases}x\ne1\\x\ge0\end{cases}}\)
\(B=\left(\frac{2\sqrt{x}+x}{x\sqrt{x}-1}-\frac{1}{\sqrt{x}-1}\right):\frac{x-1}{x+\sqrt{x}+1}\)
\(=\left(\frac{2\sqrt{x}+x}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}-\frac{x+\sqrt{x}+1}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}\right).\frac{x+\sqrt{x}+1}{x-1}\)
\(=\left(\frac{2\sqrt{x}+x-x-\sqrt{x}-1}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}\right).\frac{x+\sqrt{x}+1}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\)
\(=\frac{\sqrt{x}-1}{\left(\sqrt{x}-1\right)}.\frac{1}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\)
\(=\frac{1}{x-1}\)
20 tháng 2 2020
sửa đề: \(\frac{\sqrt{x}}{\sqrt{x}+1}\)
đkxđ: \(\left\{{}\begin{matrix}\sqrt{x}\ge0\\\sqrt{x}+1\ne0\left(lđ\right)\\x+\sqrt{x}\ne0\\\sqrt{x}-1\ne0\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x\ge0\\\sqrt{x}\left(\sqrt{x}+1\right)\ne0\\x\ne1\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x\ge0\\x\ne0\\x\ne1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x>0\\x\ne1\end{matrix}\right.\)
ta có: \(\left(\frac{\sqrt{x}}{\sqrt{x}+1}-\frac{1}{x+\sqrt{x}}\right).\frac{1}{\sqrt{x}-1}\)
=\(\left[\frac{x}{\sqrt{x}\left(\sqrt{x}+1\right)}-\frac{1}{\sqrt{x}\left(\sqrt{x}+1\right)}\right].\frac{1}{\sqrt{x}-1}\)
=\(\frac{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}{\sqrt{x}\left(\sqrt{x}+1\right)}.\frac{1}{\sqrt{x}-1}\) =\(\frac{1}{\sqrt{x}}\)
vậy...
ĐKXĐ: \(\left\{{}\begin{matrix}x\ge0\\\sqrt{x}-1\ne0\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}x\ge0\\x\ne1\end{matrix}\right.\)