N=\(\frac{2\sqrt{x}-9}{x-5\sqrt{x}+6}+\frac{2\sqrt{x}+1}{\sqrt{x}-3}+\frac{\sqrt{x}+3}{2-\sqrt{x}}\)
= \(\frac{2\sqrt{x}-9}{x-2\sqrt{x}-3\sqrt{x}+6}+\frac{2\sqrt{x}+1}{\sqrt{x}-3}-\frac{\sqrt{x}+3}{\sqrt{x}-2}\)
= \(\frac{2\sqrt{x}-9}{\sqrt{x}\left(\sqrt{x}-2\right)-3\left(\sqrt{x}-2\right)}+\frac{2\sqrt{x}+1}{\left(\sqrt{x}-3\right)}-\frac{\sqrt{x}+3}{\sqrt{x}-2}\)
= \(\frac{2\sqrt{x}-9}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}+\frac{\left(2\sqrt{x}+1\right)\left(\sqrt{x}-2\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}-\frac{\left(\sqrt{x}+3\right)\left(\sqrt{x}-2\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}\)
ĐKXĐ : x ≠ 4 ; x ≠ 9
Rút gọn :
=\(\frac{2\sqrt{x}-9}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}+\frac{\left(2\sqrt{x}+1\right)\left(\sqrt{x}-2\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}-\frac{\left(\sqrt{x}+3\right)\left(\sqrt{x}-2\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}\)
= \(\frac{2\sqrt{x}-9+\left(\sqrt{x}-2\right)\left(2\sqrt{x}+1-\sqrt{x}-3\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}\)
=\(\frac{2\sqrt{x}-9+\left(\sqrt{x}-2\right)\left(\sqrt{x}-2\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}\)
= \(\frac{2\sqrt{x}-9+x-2\sqrt{x}+4}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}\)
= \(\frac{x-5}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}\)
Để N =5 thì :
<=> \(\frac{x-5}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}\) =5
<=> x-5 = \(\left(5\sqrt{x}-10\right)\left(\sqrt{x}-3\right)\)
<=> x-5 = 5x - \(15\sqrt{x}\) - \(10\sqrt{x}\) +30
<=> x-5x-25\(\sqrt{x}\) =35
a) \(\sqrt{x}\ne3;\sqrt{x}\ne2\Rightarrow x\ne4;x\ne9\)
\(N=\frac{2\sqrt{x}-9}{x-5\sqrt{x}+6}+\frac{2\sqrt{x}+1}{\sqrt{x}-3}+\frac{\sqrt{x}+3}{2-\sqrt{x}}\)
\(\Leftrightarrow N=\frac{2\sqrt{x}-9}{\left(\sqrt{x}-3\right)\left(\sqrt{x}-2\right)}+\frac{2\sqrt{x}+1}{\sqrt{x}-3}-\frac{\sqrt{x}+3}{\sqrt{x}-2}\)
\(\Leftrightarrow N=\frac{2\sqrt{x}-9+2x-3\sqrt{x}-2-x+9}{\left(\sqrt{x}-3\right)\left(\sqrt{x}-2\right)}\)
\(\Rightarrow N=\frac{x-\sqrt{x}-2}{\left(\sqrt{x}-3\right)\left(\sqrt{x}-2\right)}=\frac{\left(\sqrt{x}-2\right)\left(\sqrt{x}+1\right)}{\left(\sqrt{x}-3\right)\left(\sqrt{x}-2\right)}=\frac{\sqrt{x}+1}{\sqrt{x}-3}\)
b) \(N=5\Leftrightarrow\frac{\sqrt{x}+1}{\sqrt{x}-3}=5\)
\(\Leftrightarrow\sqrt{x}+1=5\sqrt{x}-15\Leftrightarrow4\sqrt{x}=16\)
\(\Leftrightarrow\sqrt{x}=4\Rightarrow x=16\) (thỏa mãn)
c) \(N=\frac{\sqrt{x}+1}{\sqrt{x}-5}=\frac{\sqrt{x}-5+6}{\sqrt{x}-5}=1+\frac{6}{\sqrt{x}-5}\)
để N \(\in\) Z thì \(\left(\sqrt{x}-5\right)\inƯ\left(6\right)=\left\{\pm1;\pm2;\pm3;\pm6\right\}\)
| \(\sqrt{x}-5\) | 1 | -1 | 2 | -2 | 3 | -3 | 6 | -6 |
| x | 36 | 16 | 49 | 9 | 64 | 4 | 121 | loại |
a. ĐKXĐ: \(\left\{{}\begin{matrix}x\ge0\\x\ne4\\x\ne9\end{matrix}\right.\)
\(N=\frac{2\sqrt{x}-9}{\left(\sqrt{x}-3\right)\left(\sqrt{x}-2\right)}+\frac{\left(2\sqrt{x}+1\right)\left(\sqrt{x}-2\right)}{\left(\sqrt{x}-3\right)\left(\sqrt{x}-2\right)}-\frac{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}{\left(\sqrt{x}-3\right)\left(\sqrt{x}-2\right)}\\ =\frac{2\sqrt{x}-9+2x-4\sqrt{x}+\sqrt{x}-2-x+9}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}\\ =\frac{x-\sqrt{x}-2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}\\ =\frac{\left(\sqrt{x}-2\right)\left(\sqrt{x}+1\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}\\ =\frac{\sqrt{x}+1}{\sqrt{x}-3}\)
b.
\(N=\frac{\sqrt{x}+1}{\sqrt{x}-3}=5\\ \Leftrightarrow\sqrt{x}+1=5\left(\sqrt{x}-3\right)\\ \Leftrightarrow\sqrt{x}+1=5\sqrt{x}-15\\ \Leftrightarrow4\sqrt{x}=16\\ \Leftrightarrow\sqrt{x}=4\Leftrightarrow x=16\left(tm\right)\)
c.
\(N=\frac{\sqrt{x}+1}{\sqrt{x}-3}=\frac{\sqrt{x}-3+4}{\sqrt{x}-3}=1+\frac{4}{\sqrt{x}-3}\)
Để \(N\in Z\Leftrightarrow4⋮\sqrt{x}-3\Leftrightarrow\sqrt{x}-3\inƯ\left(4\right)\)
Ta có bảng sau:
| \(\sqrt{x}-3\) | 1 | -1 | 2 | -2 | 4 | -4 |
| \(\sqrt{x}\) | 4 | 2 | 5 | 1 | 7 | -1 |
| \(x\) | 16(tm) | 4(loại) | 25(tm) | 1(tm) | 49(tm) | loại |
Vậy......
