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a) ĐKXĐ : \(\hept{\begin{cases}x\ge0\\x\ne1\\x\ne9\end{cases}}\)
b) \(P=\left(\frac{2\sqrt{x}}{\sqrt{x}+3}+\frac{\sqrt{x}}{\sqrt{x}-3}-\frac{3x-3}{x-9}\right):\left(\frac{2\sqrt{x}-2}{\sqrt{x}+3}\right)\)
\(=\frac{2\sqrt{x}\left(\sqrt{x}-3\right)+\sqrt{x}\left(\sqrt{x}+3\right)-3x+3}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}.\frac{\sqrt{x}+3}{2\left(\sqrt{x}-1\right)}=\frac{-3\left(\sqrt{x}-1\right)}{2\left(\sqrt{x}-3\right)\left(\sqrt{x}-1\right)}=-\frac{3}{2\left(\sqrt{x}-3\right)}\)c) Để P nguyên thì \(2\left(\sqrt{x}-3\right)\in\left\{-3;-1;1;3\right\}\)=> x thuộc rỗng.
a) ĐK: \(x-9\ne0\Leftrightarrow\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)\ne0\)
Vì \(\sqrt{x}\ge0\Rightarrow\sqrt{x}+3>0\)
Nên \(\sqrt{x}-3\ne0\Leftrightarrow x\ne9\)
b) \(P=\left[\frac{2\sqrt{x}\left(\sqrt{x}-3\right)+\sqrt{x}\left(\sqrt{x}+3\right)-\left(3x+3\right)}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}\right]:\left(\frac{2\sqrt{x}-2-\left(\sqrt{x}-3\right)}{\sqrt{x}-3}\right)\)
\(=\left[\frac{2x-6\sqrt{x}+x+3\sqrt{x}-3x-3}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}\right]:\left(\frac{2\sqrt{x}-2-\sqrt{x}+3}{\sqrt{x}-3}\right)\)
\(=\left[\frac{-3\sqrt{x}-3}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}\right]\left(\frac{\sqrt{x}-3}{\sqrt{x}+1}\right)\)
\(=\left(\frac{-3\left(\sqrt{x}+1\right)}{\sqrt{x}+3}\right)\left(\frac{1}{\sqrt{x}+1}\right)\)
\(=\frac{-3}{\sqrt{x}+3}\)
c) Ta có: \(\sqrt{x}+3\ge3\)
\(\Rightarrow\frac{3}{\sqrt{x}+3}\le\frac{3}{3}=1\)
\(\Rightarrow\frac{-3}{\sqrt{x}+3}\ge-1\)
Dấu "=" xảy ra khi \(x=0\)
Vậy \(P_{min}=-1\) khi \(x=0\)
d) \(\frac{-3}{\sqrt{x}+3}< \frac{-1}{3}\)
\(\Leftrightarrow-\left(\sqrt{x}+3\right)< -9\)
\(\Leftrightarrow-\sqrt{x}< -6\)
\(\Leftrightarrow\sqrt{x}>6\)
\(\Leftrightarrow x>36\)
e) Thế \(x=3-2\sqrt{2}\) vào P ta được:
\(\frac{-3}{\sqrt{3-2\sqrt{2}}+3}=\frac{-3}{\sqrt{2}-1+3}=\frac{-3}{\sqrt{2}+2}=\frac{-3\left(\sqrt{2}-2\right)}{\left(\sqrt{2}+2\right)\left(\sqrt{2}-2\right)}=\frac{6-3\sqrt{2}}{-2}=\frac{3\sqrt{2}-6}{2}\)
f) \(P=\frac{-3}{\sqrt{x}+3}=-2\Leftrightarrow\sqrt{x}+3=6\Leftrightarrow\sqrt{x}=3\Leftrightarrow x=9\)
ĐKXĐ : \(x\ge1;x\ne2;x\ne3\)
\(P=\left[\frac{\sqrt{x}+\sqrt{x-1}}{1}-\frac{\left(x-3\right)\left(\sqrt{x-1}+\sqrt{2}\right)}{x-3}\right].\frac{2\sqrt{x}-\sqrt{x}-\sqrt{2}}{\sqrt{x}\left(\sqrt{2}-\sqrt{x}\right)}\)
\(P=\left(\sqrt{x}-\sqrt{2}\right).\frac{\left(\sqrt{x}-\sqrt{2}\right)}{\sqrt{x}\left(\sqrt{2}-\sqrt{x}\right)}=\frac{\sqrt{2}-\sqrt{x}}{\sqrt{x}}\)
\(x=3-2\sqrt{2}=\left(\sqrt{2}-1\right)^2\Rightarrow\sqrt{x}=\sqrt{2}-1\)
\(P=\frac{\sqrt{2}-\left(\sqrt{2}-1\right)}{\sqrt{2}-1}=\frac{1}{\sqrt{2}-1}=\sqrt{2}+1\)
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 \(B=\frac{2}{3-2+2\sqrt{3}}=\frac{2}{1+2\sqrt{3}}\)
1: ĐKXĐ: \(\left\{{}\begin{matrix}x\ge1\\x\ne3\end{matrix}\right.\)
2: Ta có: \(P=\dfrac{x-3}{\sqrt{x-1}-2}\)
\(=\dfrac{x-1-2}{\sqrt{x-1}-2}\)
\(=\sqrt{x-1}+2\)