помогите решить неравенство (пошагово). Created by CubossesPlayer. algebra-ru

x² - 6x + 7 = 0D = (-6)² - 4 * 7 = 36 - 28 = 8 = (2√2)² - + - +___________[-1]___________(3√2)___________(3+√2)____//////////////////////// /////////////////////////////////Ответ : x ∈ (- ∞ ; - 1] ∪ (3 - √2 ; 3 + √2)

помогите решить неравенство (пошагово)...

CubossesPlayer @CubossesPlayer

July 2022 1 6 Report

помогите решить неравенство (пошагово)

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$\frac{x^{3}+x^{2}+3x+3}{x^{2}-6x+7}\leq0\\\\\frac{(x^{3}+x^{2})+(3x+3)}{x^{2}-6x+7}\leq0$

x² - 6x + 7 = 0

D = (-6)² - 4 * 7 = 36 - 28 = 8 = (2√2)²

$x_{1}=\frac{6-2\sqrt{2}}{2}=3-\sqrt{2}\\\\x_{2}=\frac{6+2\sqrt{2}}{2}=3+\sqrt{2}\\\\\frac{x^{2}(x+1)+3(x+1)}{(x-3+\sqrt{2})(x-3-\sqrt{2})} \leq0\\\\\frac{(x+1)(x^{2}+3)}{(x-3+\sqrt{2})(x-3-\sqrt{2})}\leq0|:(x^{2}+3)>0\\\\(x+1)(x-3+\sqrt{2})(x-3-\sqrt{2})\leq0$

- + - +

___________[-1]___________(3√2)___________(3+√2)____

//////////////////////// /////////////////////////////////

Ответ : x ∈ (- ∞ ; - 1] ∪ (3 - √2 ; 3 + √2)

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Answers & Comments

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pomogite pozhalujsta reshit neravenstvo na foto190ab736718bf5b764398bb0f5ac0d17 760

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