[Toán 8] Rút gọn

I

icy_tears

a/ ĐKXĐ: $x$ khác $-3 ; 3 ; 1$
Ta có:
$x^3 - x^2 - 9x + 9$

$= (x^2 - 9)(x - 1)$


$x^5 + x^4 - 8x^3 - 12x^2 - 9x + 27$

$= x^5 - x^4 + 2x^4 - 2x^3 - 6x^3 + 6x^2 - 18x^2 + 18x - 27x + 27$

$= (x^4 + 2x^3 - 6x^2 - 18x - 27)(x - 1)$

$= (x^4 - 9x^2 + 2x^3 - 18x + 3x^2 - 27)(x - 1)$

$= (x^2 + 2x + 3)(x^2 - 9)(x - 1)$


\Rightarrow $\frac{x^3 - x^2 - 9x + 9}{x^5 + x^4 - 8x^3 - 12x^2 - 9x + 27}$

$= \frac{1}{x^2 + 2x + 3}$
 
Y

youaremysoul

$A = \dfrac{1}{x^2 + 2x +3}$
Amax \Leftrightarrow $\dfrac{1}{x^2 + 2x +3}$max
\Leftrightarrow $(x^2 + 2x +3)min$
ta có $x^2 + 2x + 3 = (x +1)^2 + 2$
vì $(x +1)^2$ \geq 0 \Rightarrow $(x +1)^2 + 2$ \geq 2
\Rightarrow $(x^2 + 2x +3)min$ =2 \Leftrightarrow x = -1
\Rightarrow $Amax = \dfrac{1}{2}$ \Leftrightarrow x = -1
 
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