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a: \(P=\left(\dfrac{2+\sqrt{x}-1}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\right):\dfrac{\sqrt{x}+1-\sqrt{x}}{\sqrt{x}+1}\)
\(=\dfrac{1}{\sqrt{x}-1}\cdot\dfrac{\sqrt{x}+1}{1}=\dfrac{\sqrt{x}+1}{\sqrt{x}-1}\)
b: Để P nguyên thì \(\sqrt{x}+1⋮\sqrt{x}-1\)
\(\Leftrightarrow\sqrt{x}-1\in\left\{-1;1;2\right\}\)
hay \(x\in\left\{0;4;9\right\}\)
a, ĐK: \(x>0;x\ne1\)
\(P=\left(\dfrac{x\sqrt{x}-1}{x-\sqrt{x}}-\dfrac{x\sqrt{x}+1}{x+\sqrt{x}}\right):\dfrac{2\left(x-2\sqrt{x}+1\right)}{x-1}\)
\(=\left[\dfrac{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}{\sqrt{x}\left(\sqrt{x}-1\right)}-\dfrac{\left(\sqrt{x}+1\right)\left(x-\sqrt{x}+1\right)}{\sqrt{x}\left(\sqrt{x}+1\right)}\right]:\dfrac{2\left(\sqrt{x}-1\right)^2}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\)
\(=\left(\dfrac{x+\sqrt{x}+1}{\sqrt{x}}-\dfrac{x-\sqrt{x}+1}{\sqrt{x}}\right).\dfrac{\sqrt{x}+1}{2\left(\sqrt{x}-1\right)}\)
\(=\dfrac{2\sqrt{x}}{\sqrt{x}}.\dfrac{\sqrt{x}+1}{2\left(\sqrt{x}-1\right)}\)
\(=\dfrac{\sqrt{x}+1}{\sqrt{x}-1}\)
a: Ta có: \(P=\left(\dfrac{x\sqrt{x}-1}{x-\sqrt{x}}-\dfrac{x\sqrt{x}+1}{x+\sqrt{x}}\right):\dfrac{2\left(x-2\sqrt{x}+1\right)}{x-1}\)
\(=\dfrac{x+\sqrt{x}+1-x+\sqrt{x}-1}{\sqrt{x}}\cdot\dfrac{\sqrt{x}+1}{2\left(\sqrt{x}-1\right)}\)
\(=\dfrac{\sqrt{x}+1}{\sqrt{x}-1}\)
b: Để P nguyên thì \(\sqrt{x}+1⋮\sqrt{x}-1\)
\(\Leftrightarrow\sqrt{x}-1\in\left\{1;-1;2\right\}\)
\(\Leftrightarrow\sqrt{x}\in\left\{2;0;3\right\}\)
ha \(x\in\left\{4;9\right\}\)
a) \(P=\dfrac{x-1+4\left(\sqrt{x}+1\right)+1}{x-1}.\dfrac{x-1}{x+2\sqrt{x}}\)
\(=\dfrac{x+4\sqrt{x}+4}{x+2\sqrt{x}}=\dfrac{\left(\sqrt{x}+2\right)^2}{\sqrt{x}\left(\sqrt{x}+2\right)}=\dfrac{\sqrt{x}+2}{\sqrt{x}}\)
b) \(P=\dfrac{\sqrt{x}+2}{\sqrt{x}}=1+\dfrac{2}{\sqrt{x}}\in Z\)
Do \(\sqrt{x}>0\)
\(\Rightarrow\sqrt{x}\inƯ\left(2\right)=\left\{1;2\right\}\)
\(\Rightarrow x\in\left\{1;4\right\}\)
a) Ta có: \(A=\left(\dfrac{2}{\sqrt{x}-3}+\dfrac{2\sqrt{x}}{x-4\sqrt{x}+3}\right):\dfrac{2\left(x-2\sqrt{x}+1\right)}{\sqrt{x}-1}\)
\(=\dfrac{2\left(\sqrt{x}-1\right)+2\sqrt{x}}{\left(\sqrt{x}-3\right)\left(\sqrt{x}-1\right)}:\dfrac{2\left(\sqrt{x}-1\right)^2}{\left(\sqrt{x}-1\right)}\)
\(=\dfrac{4\sqrt{x}-2}{\left(\sqrt{x}-3\right)\left(\sqrt{x}-1\right)}\cdot\dfrac{1}{2\left(\sqrt{x}-1\right)}\)
\(=\dfrac{2\sqrt{x}-1}{\left(\sqrt{x}-3\right)\left(\sqrt{x}-1\right)^2}\)
a: Ta có: \(N=\dfrac{x^2-\sqrt{x}}{x+\sqrt{x}+1}-\dfrac{2x+\sqrt{x}}{\sqrt{x}}+\dfrac{2\left(x-1\right)}{\sqrt{x}-1}\)
\(=x-\sqrt{x}-2\sqrt{x}-1+2\sqrt{x}+2\)
\(=x-\sqrt{x}+1\)
