Question

Answer 1

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

\(\sqrt{x}+2\in\left\{2;3;6\right\}\)

\(\Rightarrow x\in\left\{0;1;16\right\}\)

Answer 2

giai cau d th ah

 

Answer 3

`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}`