Question

bài 1: tìm x ∈ Z biết :

x² + x + 1⋮ x + 1

Answer 1

Lời giải:

Ta có:

\(x^2+x+1\vdots x+1\)

\(\Leftrightarrow x(x+1)+1\vdots x+1\)

\(\Leftrightarrow 1\vdots x+1\)

Do đó $x+1$ phải là ước của $1$

\(\Rightarrow x+1\in\left\{-1;1\right\}\)

\(\Rightarrow x\in\left\{-2;0\right\}\)

Answer 2

\(x^2+x+10⋮x+1\)

Mà \(x+1⋮x+1\)

\(\Leftrightarrow\left\{{}\begin{matrix}x^2+x+1⋮x+1\\x^2+x⋮x+1\end{matrix}\right.\)

\(\Leftrightarrow1⋮x+1\)

\(\Leftrightarrow x+1\inƯ\left(1\right)\)

\(\Leftrightarrow\left[{}\begin{matrix}x+1=1\\x+1=-1\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=-2\end{matrix}\right.\)

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