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ĐK \(\hept{\begin{cases}a\ge0\\a\ne1\end{cases}}\)
a. Ta có \(P=\frac{3a+3\sqrt{a}-3}{\left(\sqrt{a}-1\right)\left(\sqrt{a}+2\right)}-\frac{\sqrt{a}-2}{\sqrt{a}-1}+\frac{1}{\sqrt{a}+2}-1\)
\(=\frac{3a+3\sqrt{a}-3-\left(\sqrt{a}-2\right)\left(\sqrt{a}+2\right)+\sqrt{a}-1-a-\sqrt{a}+2}{\left(\sqrt{a}-1\right)\left(\sqrt{a}+2\right)}\)
\(=\frac{3a+3\sqrt{a}-3-a+4+\sqrt{a}-1-a-\sqrt{a}+2}{\left(\sqrt{a}-1\right)\left(\sqrt{a}+2\right)}=\frac{a+3\sqrt{a}+2}{\left(\sqrt{a}-1\right)\left(\sqrt{a}+2\right)}\)
\(=\frac{\left(\sqrt{a}+1\right)\left(\sqrt{a}+2\right)}{\left(\sqrt{a}-1\right)\left(\sqrt{a}+2\right)}=\frac{\left(\sqrt{a}+1\right)}{\left(\sqrt{a}-1\right)}\)
b. Để \(\left|P\right|=2\Rightarrow\orbr{\begin{cases}P=2\\P=-2\end{cases}}\)
Với \(P=2\Rightarrow\sqrt{a}+1=2\sqrt{a}-2\Rightarrow\sqrt{a}=3\Rightarrow a=9\)
Với \(P=-2\Rightarrow\sqrt{a}+1=2-2\sqrt{a}\Rightarrow\sqrt{a}=\frac{1}{3}\Rightarrow a=\frac{1}{9}\)
c. Ta có \(P=\frac{\sqrt{a}+1}{\sqrt{a}-1}=1+\frac{2}{\sqrt{a}-1}\)
Để \(P\in N\Rightarrow P\in Z\Rightarrow\sqrt{a}-1\in\left\{-2;-1;1;2\right\}\)
| \(\sqrt{a}-1\) | \(-2\) | \(-1\) | \(1\) | \(2\) |
| \(\sqrt{a}\) | \(-1\) | \(0\) | \(2\) | \(3\) |
| \(a\) | \(0\) | \(4\) | \(9\) | |
| \(\left(l\right)\) | \(\left(tm\right)\) | \(\left(tm\right)\) | \(\left(tm\right)\) |
Vậy \(x\in\left\{0;4;9\right\}\)thì \(P\in N\)
a)ĐKXĐ: \(\left\{{}\begin{matrix}x\ge0\\x\ne1\end{matrix}\right.\)
Ta có: \(A=\frac{15\sqrt{x}-11}{x+2\sqrt{x}-3}+\frac{3\sqrt{x}-2}{1-\sqrt{x}}-\frac{2\sqrt{x}+3}{\sqrt{x}+3}\)
\(=\frac{15\sqrt{x}-11}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-1\right)}-\frac{\left(3\sqrt{x}-2\right)\left(\sqrt{x}+3\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+3\right)}-\frac{\left(2\sqrt{x}+3\right)\left(\sqrt{x}-1\right)}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-1\right)}\)
\(=\frac{15\sqrt{x}-11-\left(3x+9\sqrt{x}-2\sqrt{x}-6\right)-\left(2x-2\sqrt{x}+3\sqrt{x}-3\right)}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-1\right)}\)
\(=\frac{15\sqrt{x}-11-3x-7\sqrt{x}+6-2x-\sqrt{x}+3}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-1\right)}\)
\(=\frac{-5x+7\sqrt{x}-2}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-1\right)}\)
\(=\frac{-5x+5\sqrt{x}+2\sqrt{x}-2}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-1\right)}\)
\(=\frac{-5\sqrt{x}\left(\sqrt{x}-1\right)+2\left(\sqrt{x}-1\right)}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-1\right)}\)
\(=\frac{\left(\sqrt{x}-1\right)\left(-5\sqrt{x}+2\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+3\right)}\)
\(=\frac{-5\sqrt{x}+2}{\sqrt{x}+3}\)
Ta có: \(A-\frac{2}{3}=\frac{-5\sqrt{x}+2}{\sqrt{x}+3}-\frac{2}{3}\)
\(=\frac{3\left(-5\sqrt{x}+2\right)}{3\left(\sqrt{x}+3\right)}-\frac{2\left(\sqrt{x}+3\right)}{3\left(\sqrt{x}+3\right)}\)
\(=\frac{-15\sqrt{x}+6-2\sqrt{x}-6}{3\left(\sqrt{x}+3\right)}\)
\(=\frac{-17\sqrt{x}}{3\left(\sqrt{x}+3\right)}\)
\(=\frac{-17\sqrt{x}-51+51}{3\left(\sqrt{x}+3\right)}\)
\(=\frac{-17}{3}+\frac{17}{\sqrt{x}+3}\)
Ta có: \(\sqrt{x}+3\ge3\forall x\) thỏa mãn ĐKXĐ
\(\Rightarrow\frac{17}{\sqrt{x}+3}\le\frac{17}{3}\forall x\) thỏa mãn ĐKXĐ
\(\Rightarrow\frac{17}{\sqrt{x}+3}-\frac{17}{3}\le\frac{17}{3}-\frac{17}{3}=0\forall x\) thỏa mãn ĐKXĐ
\(\Rightarrow A-\frac{2}{3}\le0\forall x\) thỏa mãn ĐKXĐ
nên \(A\le\frac{2}{3}\)(đpcm)
c) Ta có: \(C=\left(\frac{a\sqrt{a}+b\sqrt{b}}{\sqrt{a}+\sqrt{b}}-\sqrt{ab}\right):\left(a-b\right)+\frac{2\sqrt{b}}{\sqrt{a}+\sqrt{b}}\)
\(=\left(\frac{\left(\sqrt{a}+\sqrt{b}\right)\left(a-\sqrt{ab}+b\right)}{\sqrt{a}+\sqrt{b}}-\sqrt{ab}\right):\left(a-b\right)+\frac{2\sqrt{b}}{\sqrt{a}+\sqrt{b}}\)
\(=\frac{a-2\sqrt{ab}+b}{a-b}+\frac{2\sqrt{b}}{\sqrt{a}+\sqrt{b}}\)
\(=\frac{\left(\sqrt{a}-\sqrt{b}\right)^2}{\left(\sqrt{a}-\sqrt{b}\right)\left(\sqrt{a}+\sqrt{b}\right)}+\frac{2\sqrt{b}}{\sqrt{a}+\sqrt{b}}\)
\(=\frac{\sqrt{a}-\sqrt{b}+2\sqrt{b}}{\sqrt{a}+\sqrt{b}}\)
\(=\frac{\sqrt{a}+\sqrt{b}}{\sqrt{a}+\sqrt{b}}=1\)
Vậy: Giá trị của C không phụ thuộc vào a,b(đpcm)
ĐKXĐ: ...
\(P=\frac{3a+3\sqrt{a}-3}{\left(\sqrt{a}-1\right)\left(\sqrt{a}+2\right)}-\frac{\left(\sqrt{a}-2\right)\left(\sqrt{a}+2\right)}{\left(\sqrt{a}-1\right)\left(\sqrt{a}+2\right)}+\frac{\sqrt{a}-1}{\left(\sqrt{a}-1\right)\left(\sqrt{a}+2\right)}-\frac{a+\sqrt{a}-2}{\left(\sqrt{a}-1\right)\left(\sqrt{a}+2\right)}\)
\(=\frac{3a+3\sqrt{a}-3-a+4+\sqrt{a}-1-a-\sqrt{a}+2}{\left(\sqrt{a}-1\right)\left(\sqrt{a}+2\right)}\)
\(=\frac{a+3\sqrt{a}+2}{\left(\sqrt{a}-1\right)\left(\sqrt{a}+2\right)}=\frac{\left(\sqrt{a}+1\right)\left(\sqrt{a}+2\right)}{\left(\sqrt{a}-1\right)\left(\sqrt{a}+2\right)}=\frac{\sqrt{a}+1}{\sqrt{a}-1}\)
\(\left|P\right|=1\Rightarrow\left[{}\begin{matrix}P=1\\P=-1\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}\frac{\sqrt{a}+1}{\sqrt{a}-1}=1\\\frac{\sqrt{a}+1}{\sqrt{a}-1}=-1\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}\sqrt{a}+1=\sqrt{a}-1\left(vn\right)\\\sqrt{a}+1=-\sqrt{a}+1\end{matrix}\right.\) \(\Rightarrow a=0\)
\(P=\frac{\sqrt{a}-1+2}{\sqrt{a}-1}=1+\frac{2}{\sqrt{a}-1}\)
\(P\in N\Rightarrow\sqrt{a}-1=Ư\left(2\right)=\left\{-2;-1;1;2\right\}\)
\(\Rightarrow\sqrt{a}=\left\{-1\left(l\right);0;2;3\right\}\)
\(\Rightarrow a=\left\{0;4;9\right\}\)
Thay vào P chỉ thấy \(a=\left\{4;9\right\}\) thỏa mãn