a) ĐKXĐ: \(x\ne4,x\ge0\)
b) \(P=\dfrac{\sqrt{x}+1}{\sqrt{x}-2}+\dfrac{2\sqrt{x}}{\sqrt{x}+2}+\dfrac{2+5\sqrt{x}}{4-x}\)
\(P=\dfrac{\sqrt{x}+1}{\sqrt{x}-2}+\dfrac{2\sqrt{x}}{\sqrt{x}+2}-\dfrac{2+5\sqrt{x}}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}\)
\(P=\dfrac{\left(\sqrt{x}+1\right)\left(\sqrt{x}+2\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}+\dfrac{2\sqrt{x}\left(\sqrt{x}-2\right)}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}-\dfrac{2+5\sqrt{x}}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}\)
\(P=\dfrac{x+\sqrt{x}+2\sqrt{x}+2+2x-4\sqrt{x}-2-5\sqrt{x}}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}\)
\(P=\dfrac{3x-6\sqrt{x}}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}\)
\(P=\dfrac{3\sqrt{x}\left(\sqrt{x}-2\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\)
\(P=\dfrac{3\sqrt{x}}{\sqrt{x}+2}\)
c) \(P=2\) khi:
\(\dfrac{3\sqrt{x}}{\sqrt{x}+2}=2\)
\(\Leftrightarrow3\sqrt{x}=2\sqrt{x}+4\)
\(\Leftrightarrow3\sqrt{x}-2\sqrt{x}=4\)
\(\Leftrightarrow\sqrt{x}=4\)
\(\Leftrightarrow x=16\left(tm\right)\)
d) P nguên khi:
\(\dfrac{3\sqrt{x}}{\sqrt{x}+2}=\dfrac{3\sqrt{x}+6-6}{\sqrt{x}+2}=\dfrac{3\left(\sqrt{x}+2\right)-6}{\sqrt{x}+2}=3-\dfrac{6}{\sqrt{x}+2}\)
Phải nguyên:
\(\Rightarrow6\) ⋮ \(\sqrt{x}+2\)
\(\Rightarrow\sqrt{x}+2\in\left\{1;-1;2;-2;3;-3;6;-6\right\}\)
Mà: \(x\ge0\) và \(x\ne4\) nên
\(\sqrt{x}+2\in\left\{2;3;6\right\}\)
\(\Rightarrow x\in\left\{0;1;16\right\}\)
`a)P` xác định `<=>{(x >= 0),(x ne 4):}`
`b)` Với `x >= 0,x ne 4` có:
`P=[\sqrt{x}+1]/[\sqrt{x}-2]+[2\sqrt{x}]/[\sqrt{x}+2]+[2+5\sqrt{x}]/[4-x]`
`P=[(\sqrt{x}+1)(\sqrt{x}+2)+2\sqrt{x}(\sqrt{x}-2)-2-5\sqrt{x}]/[(\sqrt{x}-2)(\sqrt{x}+2)]`
`P=[x+3\sqrt{x}+2+2x-4\sqrt{x}-2-5\sqrt{x}]/[(\sqrt{x}+2)(\sqrt{x}-2)]`
`P=[3x-6\sqrt{x}]/[(\sqrt{x}+2)(\sqrt{x}-2)]`
`P=[3\sqrt{x}(\sqrt{x}-2)]/[(\sqrt{x}+2)(\sqrt{x}-2)]=[3\sqrt{x}]/[\sqrt{x}+2]`
`c)` Với `x >= 0,x ne 4` có:
`P=2<=>[3\sqrt{x}]/[\sqrt{x}+2]=2`
`<=>3\sqrt{x}=2\sqrt{x}+4`
`<=>\sqrt{x}=4<=>x=16` (t/m)
`d)` Với `x >= 0,x ne 4` có:
`P=[3\sqrt{x}]/[\sqrt{x}+2]=[3\sqrt{x}+6-6]/[\sqrt{x}+2]=3-6/[\sqrt{x}+2]`
`P` nguyên `<=>3-6/[\sqrt{x}+2] in Z`
`=>6/[\sqrt{x}+2] in ZZ`
`=>\sqrt{x}+2 in Ư_6`
Mà `Ư_6 ={+-1;+-2;+-3;+-6}`
Ta có bảng:
\begin{array}{|c|c|c|}\hline \sqrt{x}+2&1&-1&2&-2&3&-3&6&-6\\\hline x&L&L&0&L&1&L&16&L \\\hline\end{array}
Mà `x >= 0,x ne 4, x in Z =>x={0;1;16}`