`a)P` có nghĩa `<=>{(x >= 0),(\sqrt{x}-1 \ne 0):}`
`<=>{(x >= 0),(\sqrt{x} \ne 1<=>x \ne 1):}`
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`b)` Với `x >= 0,x \ne 1` có:
`P=[2x+\sqrt{x}-3]/[(\sqrt{x}+2)(\sqrt{x}-1)]-[\sqrt{x}+1]/[\sqrt{x}+2]-[\sqrt{x}-2]/[\sqrt{x}-1]`
`P=[2x+\sqrt{x}-3-(\sqrt{x}+1)(\sqrt{x}-1)-(\sqrt{x}-2)(\sqrt{x}+2)]/[(\sqrt{x}+2)(\sqrt{x}-1)]`
`P=[2x+\sqrt{x}-3-x+1-x+4]/[(\sqrt{x}+2)(\sqrt{x}-1)]`
`P=[\sqrt{x}+2]/[(\sqrt{x}+2)(\sqrt{x}-1)]`
`P=1/[\sqrt{x}-1]`
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`c)` Với `x >= 0,x \ne 1` có: `P=1/[\sqrt{x}-1]`
Để `P` có gtr nguyên `<=>1/[\sqrt{x}-1] in ZZ`
`=>\sqrt{x}-1 in Ư_1`
Mà `Ư_1={+-1}`
`@\sqrt{x}-1=1<=>\sqrt{x}=2<=>x=4` (t/m)
`@\sqrt{x}-1=-1<=>\sqrt{x}=0<=>x=0` (t/m)
a, Để P có nghĩa: \(\left\{{}\begin{matrix}x\ge0\\\left(\sqrt{x}+2\right)\left(\sqrt{x}-1\right)\ne0\\\sqrt{x}+2\ne0\\\sqrt{x}-1\ne0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x\ge0\\\sqrt{x}\ne-2\\\sqrt{x}\ne1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x\ge0\\x\ne1\end{matrix}\right.\)
b,
\(P=\dfrac{2x+\sqrt{x}-3}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-1\right)}-\dfrac{\sqrt{x}+1}{\sqrt{x}+2}-\dfrac{\sqrt{x}-2}{\sqrt{x}-1}\)
\(=\dfrac{2x+\sqrt{x}-3}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-1\right)}-\dfrac{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-1\right)}-\dfrac{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+2\right)}\)
\(=\dfrac{2x+\sqrt{x}-3}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-1\right)}-\dfrac{x-1}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-1\right)}-\dfrac{x-4}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+2\right)}\)
\(=\dfrac{2x+\sqrt{x}-3-\left(x-1\right)-\left(x-4\right)}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-1\right)}\)
\(=\dfrac{2x+\sqrt{x}-3-x+1-x+4}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-1\right)}\)
\(=\dfrac{\sqrt{x}+2}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-1\right)}=\dfrac{1}{\sqrt{x}-1}\)
c, Để P có giá trị nguyên \(\Leftrightarrow\dfrac{1}{\sqrt{x}-1}\) có giá trị nguyên \(\Leftrightarrow\sqrt{x}-1\inƯ_{\left(1\right)}=\left\{1;-1\right\}\)
\(\Leftrightarrow\sqrt{x}\in\left\{2;0\right\}\)
\(\Leftrightarrow x\in\left\{4;0\right\}\)
