Cho A = $\frac{x^2}{x-1}$ Tìm x để A < 3 06/09/2021 Bởi Harper Cho A = $\frac{x^2}{x-1}$ Tìm x để A < 3
Đáp án: $\begin{array}{l}Dkxd:x \ne 1;x \ne 0\\A = \dfrac{{{x^2}}}{{x – 1}}\\A < 3\\ \Leftrightarrow \dfrac{{{x^2}}}{{x – 1}} < 3\\ \Leftrightarrow \dfrac{{{x^2}}}{{x – 1}} – 3 < 0\\ \Leftrightarrow \dfrac{{{x^2} – 3\left( {x – 1} \right)}}{{x – 1}} < 0\\ \Leftrightarrow \dfrac{{{x^2} – 3x + 3}}{{x – 1}} < 0\\Xet:{x^2} – 3x + 3\\ = {x^2} – 2.x.\dfrac{3}{2} + \dfrac{9}{4} + \dfrac{3}{4}\\ = {\left( {x – \dfrac{3}{2}} \right)^2} + \dfrac{3}{4} > 0\\ \Leftrightarrow {x^2} – 3x + 3 > 0\\Khi:\dfrac{{{x^2} – 3x + 3}}{{x – 1}} < 0\\ \Leftrightarrow x – 1 < 0\\ \Leftrightarrow x < 1\\Vậy\,x < 1;x \ne 0\end{array}$ Bình luận
Đáp án:
$\begin{array}{l}
Dkxd:x \ne 1;x \ne 0\\
A = \dfrac{{{x^2}}}{{x – 1}}\\
A < 3\\
\Leftrightarrow \dfrac{{{x^2}}}{{x – 1}} < 3\\
\Leftrightarrow \dfrac{{{x^2}}}{{x – 1}} – 3 < 0\\
\Leftrightarrow \dfrac{{{x^2} – 3\left( {x – 1} \right)}}{{x – 1}} < 0\\
\Leftrightarrow \dfrac{{{x^2} – 3x + 3}}{{x – 1}} < 0\\
Xet:{x^2} – 3x + 3\\
= {x^2} – 2.x.\dfrac{3}{2} + \dfrac{9}{4} + \dfrac{3}{4}\\
= {\left( {x – \dfrac{3}{2}} \right)^2} + \dfrac{3}{4} > 0\\
\Leftrightarrow {x^2} – 3x + 3 > 0\\
Khi:\dfrac{{{x^2} – 3x + 3}}{{x – 1}} < 0\\
\Leftrightarrow x – 1 < 0\\
\Leftrightarrow x < 1\\
Vậy\,x < 1;x \ne 0
\end{array}$