Để H = \(\dfrac{5}{\sqrt{x}+2}\in Z\) thì \(5⋮\left(\sqrt{x}+2\right)\)
\(\Leftrightarrow\left(\sqrt{x}+2\right)\inƯ\left(5\right)\)
Mà \(\sqrt{x}+2\ge2\forall x\) nên \(\sqrt{x}+2=5\Leftrightarrow\sqrt{x}=3\Leftrightarrow x=9\)
Vậy x=9
\(H=\dfrac{5}{\sqrt{X}+2}\left(đk:x>0;x\ne2\right)\)
Ta có \(x>0\left(đkxđ\right)\Rightarrow\sqrt{x}>0\)
\(\Rightarrow\sqrt{x}+2>0\)
Mà 5>0
\(\Rightarrow\dfrac{5}{\sqrt{x}+2}>0\Rightarrow H>0\left(1\right)\)
Ta có x>0 (đkxđ)=> \(\sqrt{x}>0\)
\(\Rightarrow\sqrt{x}+2>2\rightarrow\dfrac{1}{\sqrt{x}+2}< \dfrac{1}{2}\Leftrightarrow\dfrac{5}{\sqrt{x}+2}< \dfrac{5}{2}\)
\(\Rightarrow H< \dfrac{5}{2}\left(2\right)\)
Từ 1 và 2 \(\Rightarrow0\le H< \dfrac{5}{2}\)
Mà H \(\in\) Z
=> H\(\in\)\(\left\{1;2\right\}\)
*H=1
\(\dfrac{5}{\sqrt{x}+2}=1\)
\(\sqrt{x}+2=5\)
<=>\(\sqrt{x}=3\)
<=> x=9 (t/m)
*H=2
\(\Rightarrow\dfrac{5}{\sqrt{x}+2}=2\)
=>\(2\left(\sqrt{x}+2\right)=5.1\)
\(\Leftrightarrow2\sqrt{x}+4=5\)
\(\Leftrightarrow2\sqrt{x}=1\)
\(\Leftrightarrow x=\dfrac{1}{4}\left(tmđk\right)\)
Vậy để H thuộc Z <=> x\(\in\left\{\dfrac{1}{4};9\right\}\)
chúc bạn học tót
Để H ∈ Z
\(\Rightarrow\sqrt{x}+2\inƯ\left(5\right)=\left\{1;-1;5;-5\right\}\)
| \(\sqrt{x}+2\) | 1 | -1 | 5 | -5 |
| \(x\) | ∅ | ∅ | 9 | ∅ |
Vậy \(x=9\) thì H∈Z
