Cho P = $\frac{-\sqrt{x}}{\sqrt{x}+1}$ Tìm x để P nguyên 10/08/2021 Bởi Audrey Cho P = $\frac{-\sqrt{x}}{\sqrt{x}+1}$ Tìm x để P nguyên
\(\dfrac{-\sqrt x}{\sqrt x+1}(x≥0)\\=\dfrac{-\sqrt x-1+1}{\sqrt x+1}\\=\dfrac{-(\sqrt x+1)+1}{\sqrt x+1}\\=-1+\dfrac{1}{\sqrt x+1}∈\mathbb Z\\→1\vdots \sqrt x+1\\→\sqrt x+1\in Ư(1)=\{±1\}\\→\sqrt x=0(vì\,\,\sqrt x+1>0∀x)\\→x=0\) Vậy \(x=0\) thì \(P∈\Bbb Z\) Bình luận
Đáp án:
Giải thích các bước giả
\(\dfrac{-\sqrt x}{\sqrt x+1}(x≥0)\\=\dfrac{-\sqrt x-1+1}{\sqrt x+1}\\=\dfrac{-(\sqrt x+1)+1}{\sqrt x+1}\\=-1+\dfrac{1}{\sqrt x+1}∈\mathbb Z\\→1\vdots \sqrt x+1\\→\sqrt x+1\in Ư(1)=\{±1\}\\→\sqrt x=0(vì\,\,\sqrt x+1>0∀x)\\→x=0\)
Vậy \(x=0\) thì \(P∈\Bbb Z\)