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\(a,A=\frac{2}{\sqrt{x}-3}+\frac{2\sqrt{x}}{x-4\sqrt{x}+3}+\frac{\sqrt{x}}{\sqrt{x}-1}\)
\(A=\frac{2\sqrt{x}-2+2\sqrt{x}+x-3\sqrt{x}}{\left(\sqrt{x}-1\right)\left(\sqrt{x}-3\right)}\)
\(A=\frac{x+\sqrt{x}-2}{\left(\sqrt{x}-1\right)\left(\sqrt{x}-3\right)}\)
\(A=\frac{x-\sqrt{x}+2\sqrt{x}-2}{\left(\sqrt{x}-1\right)\left(\sqrt{x}-3\right)}\)
\(A=\frac{\left(\sqrt{x}-1\right)\left(\sqrt{x}+2\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}-3\right)}\)
\(A=\frac{\sqrt{x}+2}{\sqrt{x}-3}\)
\(b,A=\frac{\sqrt{x}-3+5}{\sqrt{x}-3}=1+\frac{5}{\sqrt{x}-3}\)
để A nguyên \(5⋮\sqrt{x}-3\)
lập bảng ra đc
\(x=\left\{2\right\}\)
\(1,\frac{\sqrt{x}+1}{\sqrt{x}-3}=\frac{\sqrt{x}-3+4}{\sqrt{x}-3}=1+\frac{4}{\sqrt{x}-3}\)
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
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 => \(x\in\left\{1,25\right\}\)
a) ĐKXĐ: \(\hept{\begin{cases}x-9\ne0\\\sqrt{x}\ge0\\\sqrt{x}\ne0\end{cases}\Leftrightarrow\hept{\begin{cases}x\ne9\\x\ge0\\x\ne0\end{cases}\Leftrightarrow}\hept{\begin{cases}x\ne9\\x>0\end{cases}}}\)
\(A=\left(\frac{x+3}{x-9}+\frac{1}{\sqrt{x}+3}\right):\frac{\sqrt{x}}{\sqrt{x}-3}\)
\(\Leftrightarrow A=\left(\frac{x+3}{x-9}+\frac{\sqrt{x}-3}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}\right).\frac{\sqrt{x}-3}{\sqrt{x}}\)
\(\Leftrightarrow A=\left(\frac{x+3}{x-9}+\frac{\sqrt{x}-3}{x-9}\right).\frac{\sqrt{x}-3}{\sqrt{x}}\)
\(\Leftrightarrow A=\frac{x+\sqrt{x}}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}.\frac{\sqrt{x}-3}{\sqrt{x}}\)
\(\Leftrightarrow A=\frac{\sqrt{x}\left(\sqrt{x}+1\right)}{\sqrt{x}+3}.\frac{1}{\sqrt{x}}=\frac{\sqrt{x}+1}{\sqrt{x}+3}=\frac{\left(\sqrt{x}+1\right)\left(\sqrt{x}-3\right)}{x-9}\)
b) \(x=\sqrt{6+4\sqrt{2}}-\sqrt{3+2\sqrt{2}}\)
\(\Leftrightarrow x=\sqrt{4+4\sqrt{2}+2}-\sqrt{2+2\sqrt{2}+1}\)
\(\Leftrightarrow x=\sqrt{\left(2+\sqrt{2}\right)^2}-\sqrt{\left(\sqrt{2}+1\right)^2}\)
\(\Leftrightarrow x=\left|2+\sqrt{2}\right|-\left|\sqrt{2}+1\right|\)
\(\Leftrightarrow x=2+\sqrt{2}-\sqrt{2}-1=1\left(TM\right)\)
Vậy với x= 1 thì giá trị của biểu thức \(A=\frac{\left(1+1\right)\left(1-3\right)}{1-9}=\frac{2.\left(-2\right)}{-8}=\frac{-4}{-8}=\frac{1}{2}\)
c)
Ta có :
\(\frac{x-9}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-3\right)}=\frac{\sqrt{x}+3}{\sqrt{x}+1}=1+\frac{2}{\sqrt{x}+1}\)
+) \(\frac{1}{A}\)nguyên
\(\Leftrightarrow1+\frac{2}{\sqrt{x}+1}\)nguyên
\(\Leftrightarrow\sqrt{x}+1\inƯ\left(2\right)\)
\(\Leftrightarrow x=1\)
Vậy ..............
ĐKXĐ: \(x\ge0;x\ne\left\{4;9\right\}\)
\(A=\frac{2\sqrt{x}-9}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}+\frac{\left(2\sqrt{x}+1\right)\left(\sqrt{x}-2\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}-\frac{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}\)
\(=\frac{2\sqrt{x}-9+2x-3\sqrt{x}-2-x+9}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}=\frac{x-\sqrt{x}-2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}=\frac{\left(\sqrt{x}+1\right)\left(\sqrt{x}-2\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}=\frac{\sqrt{x}+1}{\sqrt{x}-3}\)
\(x=7-4\sqrt{3}=\left(2-\sqrt{3}\right)^2\Rightarrow\sqrt{x}=2-\sqrt{3}\)
\(\Rightarrow A=\frac{2-\sqrt{3}+1}{2-\sqrt{3}-3}=3-2\sqrt{3}\)
\(A=\frac{\sqrt{x}-3+4}{\sqrt{x}-3}=1+\frac{4}{\sqrt{x}-3}\)
Để A nguyên \(\Rightarrow\sqrt{x}-3=Ư\left(4\right)=\left\{-4;-2;-1;1;2;4\right\}\)
\(\Rightarrow\sqrt{x}=\left\{-1\left(ktm\right);1;2;4;5;7\right\}\)
\(\Rightarrow x=\left\{1;4\left(ktm\right);16;25;49\right\}\)