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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)\(A=\frac{\sqrt{x-1-2\sqrt{x-2}}}{\sqrt{x-2}-1}\)
\(=\frac{\sqrt{\left(x-2\right)-2\sqrt{x-2}+1}}{\sqrt{x-2}-1}\)
\(=\frac{\sqrt{\left(\sqrt{x-2}-1\right)^2}}{\sqrt{x-2}-1}\)
\(=\frac{\sqrt{x-2}-1}{\sqrt{x-2}-1}=1\)
\(a,ĐKXĐ:\hept{\begin{cases}x^2-\sqrt{x}\ne0\\x\ge0\\\sqrt{x}+1\ne0\end{cases}\Leftrightarrow}\hept{\begin{cases}x\ne1\\x>0\end{cases}}\)
\(b,A=\frac{1}{x^2-\sqrt{x}}:\frac{\sqrt{x}+1}{x\sqrt{x}+x+\sqrt{x}}\)
\(=\frac{1}{x^2-\sqrt{x}}\cdot\frac{x\sqrt{x}+x+\sqrt{x}}{\sqrt{x}+1}\)
\(=\frac{1}{\sqrt{x}\left(\sqrt{x}^3-1\right)}\cdot\frac{\sqrt{x}\left(x+\sqrt{x}+1\right)}{\sqrt{x}+1}\)
\(=\frac{1}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}\cdot\frac{\left(x+\sqrt{x}+1\right)}{\sqrt{x}+1}\)
\(=\frac{1}{x-1}\)
\(A=\frac{\sqrt{x}-3}{\sqrt{x}-2}-\frac{2\sqrt{x}-1}{\sqrt{x}-1}+\frac{x-2}{x-3\sqrt{x}+2}\)
\(A=\frac{\left(\sqrt{x}-3\right)\left(\sqrt{x}-1\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-1\right)}-\frac{\left(2\sqrt{x}-1\right)\left(\sqrt{x}-2\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-1\right)}\) \(+\frac{x-2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-1\right)}\)
\(A=\frac{x-4\sqrt{x}+3}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-1\right)}-\frac{2x-5\sqrt{x}+2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-1\right)}\) \(+\frac{x-2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-1\right)}\)
\(A=\frac{x-4\sqrt{x}+3-2x+5\sqrt{x}-2+x-2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-1\right)}\)
\(A=\frac{\sqrt{x}-1}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-1\right)}\)
\(A=\frac{1}{\sqrt{x}-2}\)
vậy \(A=\frac{1}{\sqrt{x}-2}\)
A có nghĩa khi \(\sqrt{x}-2>0\)
\(\Leftrightarrow\sqrt{x}=2\)
\(\Leftrightarrow x=4\)
vậy \(x=4\) thì A có nghĩa
b) theo ý a) \(A=\frac{1}{\sqrt{x}-2}\)
theo bài ra \(A>2\) \(\Leftrightarrow\frac{1}{\sqrt{x}-2}>2\)
\(\Leftrightarrow\frac{1}{\sqrt{x}-2}-2>0\)
\(\Leftrightarrow\frac{1}{\sqrt{x}-2}-\frac{2\left(\sqrt{x}-2\right)}{\sqrt{x}-2}>0\)
\(\Leftrightarrow\frac{1-2\sqrt{x}+4}{\sqrt{x}-2}>0\)
\(\Leftrightarrow\frac{5-2\sqrt{x}}{\sqrt{x}-2}>0\)
\(\Rightarrow\hept{\begin{cases}5-2\sqrt{x}>0\\\sqrt{x}-2>0\end{cases}}\) hoặc \(\hept{\begin{cases}5-2\sqrt{x}< 0\\\sqrt{x}-2< 0\end{cases}}\)
\(\Rightarrow\hept{\begin{cases}-2\sqrt{x}>-5\\\sqrt{x}>2\end{cases}}\) hoặc \(\hept{\begin{cases}-2\sqrt{x}< -5\\\sqrt{x}< 2\end{cases}}\)
\(\Rightarrow\hept{\begin{cases}x< \frac{25}{4}\\x>4\end{cases}}\)hoặc \(\hept{\begin{cases}x>\frac{25}{4}\\x< 4\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}4< x< \frac{25}{4}\\x\notin\varnothing\end{cases}}\)
vậy \(4< x< \frac{25}{4}\) thì \(A>2\)
1,
\(A=\left(\frac{a\sqrt{a}-1}{a-\sqrt{a}}-\frac{a\sqrt{a}+1}{a+\sqrt{a}}\right):\frac{a+2}{a-2}\left(đk:a\ne0;1;2;a\ge0\right)\)
\(=\frac{\left(a\sqrt{a}-1\right)\left(a+\sqrt{a}\right)-\left(a\sqrt{a}+1\right)\left(a-\sqrt{a}\right)}{a^2-a}.\frac{a-2}{a+2}\)
\(=\frac{a^2\sqrt{a}+a^2-a-\sqrt{a}-\left(a^2\sqrt{a}-a^2+a-\sqrt{a}\right)}{a\left(a-1\right)}.\frac{a-2}{a+2}\)
\(=\frac{2a\left(a-1\right)\left(a-2\right)}{a\left(a-1\right)\left(a+2\right)}=\frac{2\left(a-2\right)}{a+2}\)
Để \(A=1\)\(=>\frac{2a-4}{a+2}=1< =>2a-4-a-2=0< =>a=6\)
2,
a, Điều kiện xác định của phương trình là \(x\ne4;x\ge0\)
b, Ta có : \(B=\frac{2\sqrt{x}}{x-4}+\frac{1}{\sqrt{x}-2}-\frac{1}{\sqrt{x}+2}\)
\(=\frac{2\sqrt{x}}{x-4}+\frac{\sqrt{x}+2}{x-4}-\frac{\sqrt{x}-2}{x-4}\)
\(=\frac{2\sqrt{x}+2+2}{x-4}=\frac{2\left(\sqrt{x}+2\right)}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}=\frac{2}{\sqrt{x}-2}\)
c, Với \(x=3+2\sqrt{3}\)thì \(B=\frac{2}{3-2+2\sqrt{3}}=\frac{2}{1+2\sqrt{3}}\)
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 \(2\le x< 3\)