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\(1,\frac{\sqrt{x}+1}{\sqrt{x}-3}=\frac{\sqrt{x}-3+4}{\sqrt{x}-3}=1+\frac{4}{\sqrt{x}-3}\)
Để \(\frac{\sqrt{x}+1}{\sqrt{x}-3}\in Z\Rightarrow\frac{4}{\sqrt{x}-3}\in Z\)
\(\Rightarrow\sqrt{x}-3\in\left(1;4;-1;-4\right)\)
\(\Rightarrow\sqrt{x}\in\left(4;7;2;-1\right)\)
\(\Rightarrow\sqrt{x}=4\Leftrightarrow x=2\)
\(4,A=x+\sqrt{x}+1\)
\(A=\left(\sqrt{x}\right)^2+2.\frac{1}{2}.\sqrt{x}+\left(\frac{1}{2}\right)^2+\frac{3}{4}\)
\(A=\left(\sqrt{x}+\frac{1}{2}\right)^2+\frac{3}{4}\)
\(\Rightarrow A\ge\frac{3}{4}.\left(\sqrt{x}+\frac{1}{2}\right)^2\ge0\)
Dấu "=" xảy ra khi :
\(\sqrt{x}+\frac{1}{2}=0\Leftrightarrow\sqrt{x}=-\frac{1}{2}\)
Vậy Min A = 3/4 khi căn x = -1/2
a: \(A=\dfrac{\left(\sqrt{a}-\sqrt{b}\right)^2}{\sqrt{a}-\sqrt{b}}-\dfrac{\sqrt{ab}\left(\sqrt{a}+\sqrt{b}\right)}{\sqrt{ab}}\)
\(=\sqrt{a}-\sqrt{b}-\sqrt{a}-\sqrt{b}=-2\sqrt{b}\)
b: \(B=\dfrac{2\sqrt{x}-x-x-\sqrt{x}-1}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}\cdot\dfrac{x+\sqrt{x}+1}{x-1}\)
\(=\dfrac{-2x+\sqrt{x}-1}{\sqrt{x}-1}\cdot\dfrac{1}{x-1}\)
c: \(C=\dfrac{x-9-x+3\sqrt{x}}{x-9}:\left(\dfrac{3-\sqrt{x}}{\sqrt{x}-2}+\dfrac{\sqrt{x}-2}{\sqrt{x}+3}+\dfrac{x-9}{x+\sqrt{x}-6}\right)\)
\(=\dfrac{3\left(\sqrt{x}-3\right)}{x-9}:\dfrac{9-x+x-4\sqrt{x}+4+x-9}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-2\right)}\)
\(=\dfrac{3}{\sqrt{x}+3}\cdot\dfrac{\left(\sqrt{x}+3\right)\left(\sqrt{x}-2\right)}{x-4\sqrt{x}+4}\)
\(=\dfrac{3}{\sqrt{x}-2}\)
ĐKCĐ: \(x\ge0;x\ne9,x\ne4\)
\(A=\left(\frac{x-3\sqrt{x}}{x-9}-1\right):\left(\frac{9-x}{x+\sqrt{x}-6}+\frac{\sqrt{x}-3}{\sqrt{x}-2}-\frac{\sqrt{x}-2}{\sqrt{x}+3}\right)\\ \)
\(=\left(\frac{\sqrt{x}.\left(\sqrt{x}-3\right)}{\left(\sqrt{x}-3\right).\left(\sqrt{x}+3\right)}-1\right):\left(\frac{\left(3-\sqrt{x}\right).\left(3+\sqrt{x}\right)}{\left(\sqrt{x}-2\right).\left(\sqrt{x+3}\right)}+\frac{\sqrt{x}-3}{\sqrt{x}-2}-\frac{\sqrt{x}-2}{\sqrt{x}+3}\right)\)
\(=\left(\frac{\sqrt{x}}{\sqrt{x}+3}-1\right):\left(\frac{3-\sqrt{x}}{\sqrt{x}-2}+\frac{\sqrt{x}-3}{\sqrt{x}-2}-\frac{\sqrt{x}-2}{\sqrt{x}+3}\right)\)
\(=-\frac{3}{\sqrt{x}+3}:\left(-\frac{\sqrt{x}-2}{\sqrt{x}+3}\right)=-\frac{3}{\sqrt{x}+3}:\frac{-\left(\sqrt{x}-2\right)}{\sqrt{x}+3}=\frac{3}{\sqrt{x}-2}\)
b, \(A\inℤ\Leftrightarrow\frac{3}{\sqrt{x}-2}\inℤ\)
Nếu x không là số chính phương thì \(\sqrt{x}\)là số vô tỉ thì \(\sqrt{x}-2\)là số vô tỉ\(\Rightarrow A=\frac{3}{\sqrt{x}-2}\)là số vô tỉ
Nếu x là số chính phương thì \(\sqrt{x}\)là số nguyên thì \(\sqrt{x}-2\inℤ\Rightarrow\sqrt{x}-2\inƯ\left(3\right)\Rightarrow\sqrt{x}-2\in\left\{\pm1;\pm3\right\}\Rightarrow\sqrt{x}\in\left\{1;3;5\right\}\)\(\Rightarrow x\in\left\{1;9;25\right\}\)
Mà theo ĐKXĐ có x khác 9 => \(x\in\left\{1,25\right\}\)