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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: \(C=\dfrac{x-9-x+3\sqrt{x}}{x-9}:\left(\dfrac{3-\sqrt{x}}{\sqrt{x}-2}+\dfrac{\sqrt{x}-2}{\sqrt{x}+3}+\dfrac{x-9}{x+\sqrt{x}-6}\right)\)
\(=\dfrac{3\left(\sqrt{x}-3\right)}{x-9}:\dfrac{9-x+x-4\sqrt{x}+4+x-9}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-2\right)}\)
\(=\dfrac{3}{\sqrt{x}+3}\cdot\dfrac{\left(\sqrt{x}+3\right)\left(\sqrt{x}-2\right)}{x-4\sqrt{x}+4}\)
\(=\dfrac{3}{\sqrt{x}-2}\)
\(A=\frac{\left(\sqrt{x}^3-1\right)\left(x+\sqrt{x}\right)-\left(\sqrt{x}^3+1\right)\left(x-\sqrt{x}\right)}{\left(x-\sqrt{x}\right)\left(x+\sqrt{x}\right)}.\frac{x-1}{2\left(\sqrt{x}-1\right)^2}\)
\(A=\frac{\sqrt{x}^5+x^2-x-\sqrt{x}-\sqrt{x}^5+x^2-x+\sqrt{x}}{x^2-x}.\frac{x-1}{2\left(\sqrt{x}-1\right)^2}\)
\(A=\frac{2x^2-2x}{x^2-x}.\frac{x-1}{2\left(\sqrt{x}-1\right)^2}\)
\(A=\frac{2x\left(x-1\right)}{x\left(x-1\right)}.\frac{x-1}{2\left(\sqrt{x}-1\right)^2}\)
\(A=\frac{2\left(x-1\right)}{2\left(\sqrt{x}-1\right)^2}\)
\(A=\frac{x-1}{\left(\sqrt{x}-1\right)^2}\)
\(A=\frac{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}{\left(\sqrt{x}-1\right)^2}=\frac{\sqrt{x}+1}{\sqrt{x}-1}\)