[Toán 8]

I

iceghost

$A = \dfrac{x-2}{x^3-x^2-x-2} \\
= \dfrac{x-2}{x^3-2x^2+x^2-2x+x-2} \\
= \dfrac{x-2}{x^2(x-2)+x(x-2)+(x-2)} \\
= \dfrac{x-2}{(x-2)(x^2+x+1)} \\
= \dfrac{1}{x^2+x+1} \\
= \dfrac{1}{x^2+2.x.\dfrac12+\dfrac14+\dfrac34} \\
= \dfrac{1}{(x+\dfrac12)^2+\dfrac34}$
Có : $(x+\dfrac12)^2 \ge 0$
$\iff (x+\dfrac12)^2+\dfrac34 \ge \dfrac34 \\
\implies \dfrac{1}{(x+\dfrac12)^2+\dfrac34} \le \dfrac43 \\
\implies \textrm{Max}_A = \dfrac43 \iff x+\dfrac12=0 \iff x=-\dfrac12$
 
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