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a) \(ĐKXĐ:x\ne4;x\ne9\)
b) \(A=\frac{2\sqrt{x}-9}{x-5\sqrt{x}+6}-\frac{\sqrt{x}+3}{\sqrt{x}-2}-\frac{2\sqrt{x}+1}{3-\sqrt{x}}\)
\(=\frac{2\sqrt{x}-9}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}-\frac{\sqrt{x}+3}{\sqrt{x}-2}+\frac{2\sqrt{x}+1}{\sqrt{x}-3}\)
\(=\frac{2\sqrt{x}-9-\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)+\left(2\sqrt{x}+1\right)\left(\sqrt{x}-2\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}\)
\(=\frac{2\sqrt{x}-9-x+9+2x-3\sqrt{x}-2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}=\frac{-\sqrt{x}+x-2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}\)
\(=\frac{x-\sqrt{x}-2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}=\frac{\left(\sqrt{x}+1\right)\left(\sqrt{x}-2\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}=\frac{\sqrt{x}+1}{\sqrt{x}-3}\)
c) Ta có: \(A=\frac{\sqrt{x}+1}{\sqrt{x}-3}=\frac{\left(\sqrt{x}-3\right)+4}{\sqrt{x}-3}=1+\frac{4}{\sqrt{x}-3}\)
\(\Rightarrow\sqrt{x}-3\inƯ\left(4\right)=\left\{\pm1;\pm2;\pm4\right\}\) (ĐK: x thuộc Z)
| \(\sqrt{x}-3\) | 1 | -1 | 2 | -2 | 4 | -4 |
| \(\sqrt{x}\) | 4 | 2 | 5 | 1 | 7 | -1 |
| x | 2 | \(\sqrt{2}\) | \(\sqrt{5}\) | \(\sqrt{1}\) | \(\sqrt{7}\) | \(\varnothing\) |
Vậy để A thuộc Z khi x = {2;\(\sqrt{2};\sqrt{5};\sqrt{1};\sqrt{7}\) }
a) \(A=\left(\frac{x+3}{x-9}+\frac{1}{\sqrt{x}+3}\right):\frac{\sqrt{x}}{\sqrt{x}-3}\)
\(=\left[\frac{x+3+\sqrt{x}-3}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}\right]:\frac{\sqrt{x}}{\sqrt{x}-3}\)
\(=\frac{\sqrt{x}\left(\sqrt{x}+1\right)}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}.\frac{\sqrt{x}-3}{\sqrt{x}}\)
\(=\frac{\sqrt{x}+1}{\sqrt{x}+3}\)
c) để A>1/3
\(\Rightarrow\frac{\sqrt{x}+3-2}{\sqrt{x}+3}>\frac{1}{3}\)
\(\Rightarrow\frac{2}{\sqrt{x}+3}>\frac{2}{3}\)
\(\Rightarrow\sqrt{x}+3>3\)
\(\Rightarrow x>0\)
\(B=\frac{\sqrt{x}+2}{\sqrt{x}-3}-\frac{\sqrt{x}+1}{\sqrt{x}-2}+\frac{3\left(\sqrt{x}-1\right)}{x-5\sqrt{x}+6}\left(ĐKXĐ:x\ne4;x\ne9;x\ge0\right)\)
\(=\frac{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}{\left(\sqrt{x}-3\right)\left(\sqrt{x}-2\right)}-\frac{\left(\sqrt{x}+1\right)\left(\sqrt{x}-3\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}-\frac{3\left(\sqrt{x}-1\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}\)
\(=\frac{x-4-\left(x-2\sqrt{x}-3\right)-3\left(\sqrt{x}-1\right)}{\left(\sqrt{x}-3\right)\left(\sqrt{x}-2\right)}=\frac{2-\sqrt{x}}{\left(\sqrt{x}-3\right)\left(\sqrt{x}-2\right)}=\frac{1}{3-\sqrt{x}}\)
\(B< -1\)\(\Leftrightarrow\) \(\frac{1}{3-\sqrt{x}}< -1\)\(\Rightarrow\sqrt{x}-3< 1\Leftrightarrow x< 16\)
Mặt khác : Vì \(B< -1< 0\)nên \(3-\sqrt{x}< 0\Rightarrow x>9\)
Vậy để \(B< -1\)thì \(9< x< 16\)
\(2B\in Z\Leftrightarrow B\in Z\)
\(\Leftrightarrow\frac{1}{3-\sqrt{x}}\in Z\)=> \(3-\sqrt{x}\inƯ\left(1\right)\)
\(\Rightarrow3-\sqrt{x}\in\left\{-1;1\right\}\)\(\Rightarrow x\in\left\{16\right\}\)( Loại x = 4 vì không thoả mãn điều kiện)
Xin lỗi vì để bài mình ghi lộn :))
Còn lại thì ổn rồi :))
a, Với \(x\ge0;x\ne1\)
\(B=\frac{1}{\sqrt{x}-1}=2\Rightarrow2\sqrt{x}-2=1\Leftrightarrow2\sqrt{x}-3=0\Leftrightarrow x=\frac{9}{4}\)
b, Ta có : \(A.B=\frac{x+3}{\sqrt{x}+1}.\frac{1}{\sqrt{x}-1}=\frac{x+3}{x-1}=\frac{x-1+4}{x-1}=1+\frac{4}{x-1}\)
\(\Rightarrow x-1\inƯ\left(4\right)=\left\{\pm1;\pm2;\pm4\right\}\)
c, Ta có : \(A=\frac{x+3}{\sqrt{x}+1}\le3\Leftrightarrow\frac{x+3}{\sqrt{x}+1}-3\le0\)
\(\Leftrightarrow\frac{x-3\sqrt{x}}{\sqrt{x}+1}\le0\Rightarrow\sqrt{x}-3\le0\Leftrightarrow x\le9\)
Kết hợp với đk vậy 0 =< x =< 9