Question

Tìm x \(\in\) z để M = \(\dfrac{\sqrt{x}+1}{\sqrt{x}+3}\)nguyên dương

giúp mk nha

Answer 1

\(M=\dfrac{\sqrt{x}+1}{\sqrt{x}+3}=\dfrac{\sqrt{x}+3-2}{\sqrt{x}+3}=\dfrac{\sqrt{x}+3}{\sqrt{x}+3}-\dfrac{2}{\sqrt{x}+3}=1-\dfrac{2}{\sqrt{x}+3}\)Để \(M\in Z^+\) thì: \(2⋮\sqrt{x}+3\) và \(\dfrac{2}{\sqrt{x}+3}< 1\)

\(\Rightarrow\left[{}\begin{matrix}\sqrt{x}+3=2\\\sqrt{x}+3=-2\\\sqrt{x}+3=1\\\sqrt{x}+3=-1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}\sqrt{x}=-1\\\sqrt{x}=-5\\\sqrt{x}=-2\\\sqrt{x}=-4\end{matrix}\right.\left(loại\right)\)

Vậy \(x\in\varnothing\)

Answer 2

vãi. lm Toán luôn :)

Answer 3

thanks bn nhiều nha Ribi Nkok Ngok