Question

Tìm x nguyên để P nguyên.
P= \(\dfrac{\sqrt{x}+7}{\sqrt{x}+2}\)

Answer 1

ĐKXĐ: \(x\ge0\)

Ta có \(P=\dfrac{\sqrt{x}+7}{\sqrt{x}+2}=\dfrac{\sqrt{x}+2+5}{\sqrt{x}+2}=1+\dfrac{5}{\sqrt{x}+2}\) 

Để P nguyên thì \(5⋮\left(\sqrt{x}+2\right)\Rightarrow\left(\sqrt{x}+2\right)\inƯ\left(5\right)=\left\{1;-1;5;-5\right\}\)

\(\Rightarrow\left[{}\begin{matrix}\sqrt{x}+2=1\\\sqrt{x}+2=-1\\\sqrt{x}+2=5\\\sqrt{x}+2=-5\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}\sqrt{x}=-1\\\sqrt{x}=-3\\\sqrt{x}=3\\\sqrt{x}=-7\end{matrix}\right.\) \(\Rightarrow x=9\) ( thỏa mãn đk )

Vậy x thỏa mãn ycbt là x=9

 

Answer 2

Để P nguyên thì \(\sqrt{x}+7⋮\sqrt{x}+2\)

\(\Leftrightarrow\sqrt{x}+2=5\)

hay x=9

Answer 3

\(P=\dfrac{\sqrt{x}+7}{\sqrt{x}+2}=\dfrac{\sqrt{x}+2}{\sqrt{x}+2}+\dfrac{5}{\sqrt{x}+2}=1+\dfrac{5}{\sqrt{x}+2}\)

Để P nguyên thì \(\sqrt{x}+2\inƯ\left(5\right)=\left\{1;-1;5;-5\right\}\)

Do x nguyên và \(x\ge0\)

\(\Rightarrow x=\sqrt{3}\)

Answer 4

Mik ko hiểu đề lắm nhỉ