Câu hỏi của Edogawa Conan

Question

tìm x nguyên để $B=\dfrac{\sqrt{x+6}}{\sqrt{x+1}}nguyên$

Rimuru tempest · Accepted Answer

$B=\sqrt{\dfrac{x+6}{x+1}}=\sqrt{\dfrac{x+1+5}{x+1}}=\sqrt{1+\dfrac{5}{x+1}}$ (Đk $x>-1$ )

Để $\dfrac{5}{x+1}\in Z$ khi $\left(x+1\right)\inƯ\left(5\right)=\left\{-5;-1;1;5\right\}$

$\Leftrightarrow\left\{{}\begin{matrix}x+1=5\x+1=1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=4\x=0\end{matrix}\right.$

Để B nguyên khi $\dfrac{5}{x+1}=n^2-1$ $\left(n\in N\right)$

Trường hợp x=0 $\Leftrightarrow5=n^2-1\Leftrightarrow n^2=6\Leftrightarrow n=\pm\sqrt{6}\left(loại\right)$

Trường hợp x=4 $\Leftrightarrow1=n^2-1\Leftrightarrow n^2=2\Leftrightarrow n=\pm\sqrt{2}\left(loại\right)$

vậy $S=\varnothing$Câu trả lời: $B=\sqrt{\dfrac{x+6}{x+1}}=\sqrt{\dfrac{x+1+5}{x+1}}=\sqrt{1+\dfrac{5}{x+1}}$ (Đk $x>-1$ )

Để $\dfrac{5}{x+1}\in Z$ khi $\left(x+1\right)\inƯ\left(5\right)=\left\{-5;-1;1;5\right\}$

$\Leftrightarrow\left\{{}\begin{matrix}x+1=5\x+1=1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=4\x=0\end{matrix}\right.$

Để B nguyên khi $\dfrac{5}{x+1}=n^2-1$ $\left(n\in N\right)$

Trường hợp x=0 $\Leftrightarrow5=n^2-1\Leftrightarrow n^2=6\Leftrightarrow n=\pm\sqrt{6}\left(loại\right)$

Trường hợp x=4 $\Leftrightarrow1=n^2-1\Leftrightarrow n^2=2\Leftrightarrow n=\pm\sqrt{2}\left(loại\right)$

vậy $S=\varnothing$

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