Toán Lớp 8: Tìm x nguyên để biểu thức có giá trị nguyên :
$\frac{x^{3}-x^{2}}{x^{2}-2x+1}$
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Toán Lớp 8: Tìm x nguyên để biểu thức có giá trị nguyên :
$\frac{x^{3}-x^{2}}{x^{2}-2x+1}$
Comments ( 2 )
Dkxd:x \ne 1\\
A = \dfrac{{{x^3} – {x^2}}}{{{x^2} – 2x + 1}} = \dfrac{{{x^2}\left( {x – 1} \right)}}{{{{\left( {x – 1} \right)}^2}}} = \dfrac{{{x^2}}}{{x – 1}}\\
= \dfrac{{{x^2} – 1 + 1}}{{x – 1}}\\
= \dfrac{{\left( {x – 1} \right)\left( {x + 1} \right) + 1}}{{x – 1}}\\
= x + 1 + \dfrac{1}{{x – 1}}\\
A \in Z\\
\Leftrightarrow \dfrac{1}{{x – 1}} \in Z\\
\Leftrightarrow \left( {x – 1} \right) \in \left\{ { – 1;1} \right\}\\
\Leftrightarrow x \in \left\{ {0;2} \right\}\left( {tmdk} \right)\\
Vậy\,x \in \left\{ {0;2} \right\}
\end{array}$