Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
Bài 1:
a: \(B=\dfrac{\sqrt{x}+x+\sqrt{x}-x}{1-x}\cdot\dfrac{x-1}{3-\sqrt{x}}\)
\(=\dfrac{2\sqrt{x}}{\sqrt{x}-3}\)
b: Để B=-1 thì \(2\sqrt{x}=-\sqrt{x}+3\)
=>3 căn x=3
=>căn x=1
hay x=1(loại)
a) ĐKXĐ: x≠0;x≠1;x>0
b) \(A=\left(\dfrac{2\sqrt{x}+x}{x\sqrt{x}-x}-\dfrac{1}{\sqrt{x}-1}\right)\div\dfrac{x-1}{x+\sqrt{x}+1}=\left(\dfrac{\sqrt{x}\left(2+\sqrt{x}\right)}{x\left(\sqrt{x}-1\right)}-\dfrac{1}{\sqrt{x}-1}\right).\dfrac{x+\sqrt{x}+1}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}=\left(\dfrac{2+\sqrt{x}-\sqrt{x}}{\sqrt{x}\left(\sqrt{x}-1\right)}\right).\dfrac{x+\sqrt{x}+1}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}=\dfrac{2x+2\sqrt{x}+2}{\sqrt{x}\left(\sqrt{x}-1\right)^2\left(\sqrt{x}+1\right)}\)
\(a,ĐKXĐ:\hept{\begin{cases}x^2-\sqrt{x}\ne0\\x\ge0\\\sqrt{x}+1\ne0\end{cases}\Leftrightarrow}\hept{\begin{cases}x\ne1\\x>0\end{cases}}\)
\(b,A=\frac{1}{x^2-\sqrt{x}}:\frac{\sqrt{x}+1}{x\sqrt{x}+x+\sqrt{x}}\)
\(=\frac{1}{x^2-\sqrt{x}}\cdot\frac{x\sqrt{x}+x+\sqrt{x}}{\sqrt{x}+1}\)
\(=\frac{1}{\sqrt{x}\left(\sqrt{x}^3-1\right)}\cdot\frac{\sqrt{x}\left(x+\sqrt{x}+1\right)}{\sqrt{x}+1}\)
\(=\frac{1}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}\cdot\frac{\left(x+\sqrt{x}+1\right)}{\sqrt{x}+1}\)
\(=\frac{1}{x-1}\)
a/ ĐKXĐ: \(\hept{\begin{cases}x-2\ge0\\\sqrt{x-2}-1\ne0\end{cases}\Rightarrow\hept{\begin{cases}x\ge2\\x-2\ne1\end{cases}\Rightarrow}\hept{\begin{cases}x\ge2\\x\ne3\end{cases}}}\)
b/ \(A=\frac{\sqrt{x-2-2\sqrt{x-2}+1}}{\sqrt{x-2}-1}=\frac{\sqrt{\left(\sqrt{x-2}-1\right)^2}}{\sqrt{x-2}-1}=\frac{\left|\sqrt{x-2}-1\right|}{\sqrt{x-2}-1}\left(1\right)\)
+ Khi \(\sqrt{x-2}-1>0\Rightarrow x-2>1\Rightarrow x>3\) thì (1) trở thành:
\(A=\frac{\sqrt{x-2}-1}{\sqrt{x-2}-1}=1\)
+ Khi \(\sqrt{x-2}-1< 0\Rightarrow x< 3\) thì (1) trở thành:
\(A=\frac{1-\sqrt{x-2}}{\sqrt{x-2}-1}=-1\)
Vậy A = 1 khi x > 3
A = -1 khi \(2\le x< 3\)
ĐK:\(\begin{cases}x-2\ge0\\x-1-2\sqrt{x-2}\ge0\\\sqrt{x-2}-1\ne0\end{cases}\)\(\Leftrightarrow\begin{cases}x\ge2\\\left(x-2\right)-2\sqrt{x-2}+1\ge0\\\sqrt{x-2}\ne1\end{cases}\)
\(\Leftrightarrow\begin{cases}x\ge2\\\left(\sqrt{x-2}-1\right)^2\ge0\\x-2\ne1\end{cases}\)\(\Leftrightarrow\begin{cases}x\ge2\\\sqrt{x-2}\ge1\\x\ne3\end{cases}\)\(\Leftrightarrow\begin{cases}x\ge2\\x-2\ge1\\x\ne3\end{cases}\)
\(\Leftrightarrow\begin{cases}x\ge2\\x\ge3\\x\ne3\end{cases}\) \(\Leftrightarrow x>3\)
b)\(A=\frac{\sqrt{x-1-2\sqrt{x-2}}}{\sqrt{x-2}-1}\)
\(=\frac{\sqrt{\left(x-2\right)-2\sqrt{x-2}+1}}{\sqrt{x-2}-1}\)
\(=\frac{\sqrt{\left(\sqrt{x-2}-1\right)^2}}{\sqrt{x-2}-1}\)
\(=\frac{\sqrt{x-2}-1}{\sqrt{x-2}-1}=1\)
Ta có A=\(\dfrac{\sqrt{x}-2}{\sqrt{x}+1}\) với x≥ 9, x ∈ R
Để A > 0 \(\Leftrightarrow\) \(\dfrac{\sqrt{x}-2}{\sqrt{x}+1}\) > 0
\(\Leftrightarrow\) \(\left[{}\begin{matrix}\left\{{}\begin{matrix}\sqrt{x}-2>0\\\sqrt{x}+1< 0\end{matrix}\right.\\\left\{{}\begin{matrix}\sqrt{x}-2< 0\\\sqrt{x}+1>0\end{matrix}\right.\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}\sqrt{x}>2\\\sqrt{x}< -1\end{matrix}\right.\\\left\{{}\begin{matrix}\sqrt{x}< 2\\\sqrt{x}>-1\end{matrix}\right.\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x>4\\x< 1\end{matrix}\right.\\\left\{{}\begin{matrix}x< 4\\x>1\end{matrix}\right.\end{matrix}\right.\)
Kết hợp với ĐKXĐ\(\Rightarrow\) x ∈ ∅
ĐKXĐ: x≥9, x∈R
Ta có:
A= \(\left[\dfrac{1+\sqrt{x}-\sqrt{x}}{1+\sqrt{x}}\right]\):\(\left[\dfrac{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}-\dfrac{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}{\left(\sqrt{x}-3\right)\left(\sqrt{x}-2\right)}+\dfrac{\sqrt{x}+2}{x-2\sqrt{x}-3\sqrt{x}+6}\right]\)
= \(\left[\dfrac{1}{1+\sqrt{x}}\right]\):\(\left[\dfrac{x-9}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}-\dfrac{x-4}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}+\dfrac{\sqrt{x}+2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}\right]\)
=\(\left[\dfrac{1}{1+\sqrt{x}}\right]\):\(\left[\dfrac{x-9-x+4+\sqrt{x}+2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}\right]\)
=\(\left[\dfrac{1}{1+\sqrt{x}}\right]\):\(\left[\dfrac{\sqrt{x}-3}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}\right]\)
=\(\dfrac{1}{1+\sqrt{x}}\):\(\dfrac{1}{\sqrt{x}-2}\)
=\(\dfrac{\sqrt{x}-2}{\sqrt{x}+1}\)
a) A có nghĩa khi \(x>0;x\ne1\)
b)
\(A=\dfrac{1}{x^2-\sqrt{x}}:\dfrac{\sqrt{x}+1}{x\sqrt{x}+x+\sqrt{x}}\\ A=\dfrac{1}{\sqrt{x}\left(\sqrt{x}^3-1\right)}:\dfrac{\sqrt{x}+1}{\sqrt{x}\left(x+\sqrt{x}+1\right)}\\ A=\dfrac{1}{\sqrt{x}\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}:\dfrac{\sqrt{x}+1}{\sqrt{x}\left(x+\sqrt{x}+1\right)}\\ A=\dfrac{\sqrt{x}\left(x+\sqrt{x}+1\right)}{\sqrt{x}\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)\left(\sqrt{x}+1\right)}\\ A=\dfrac{1}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}=\dfrac{1}{x-1}\)
Xin lỗi nha...Chỗ giữa kia thì dấu''.''thay bằng dấu'':'' hộ mình với.