Решите уравнение: x^3-x^2-x+2=0. Created by dashamilasha2402. algebra-ru

Решите уравнение: x^3-x^2-x+2=0...

dashamilasha2402 @dashamilasha2402

August 2022 1 6 Report

Решите уравнение:

x^3-x^2-x+2=0

Answers & Comments

mathkot

Ответ:

$\boxed{ x \approx -1,21 }$

Объяснение:

$x \in \mathbb R$

$x^{3} - x^{2} - x + 2 = 0$

$x^{3} + (-1)\cdot x^{2} + (-1)\cdot x + 2 = 0$

Воспользуемся тригонометрической формулой Виета для решения кубических уравнений.

$Q = \dfrac{a^{2} - 3b}{9} = \dfrac{(-1)^{2} - 3\cdot (-1)}{9} = \dfrac{1 + 3}{9} = \dfrac{4}{9}$

$R = \dfrac{2a^{3} - 9ab + 27c}{54} = \dfrac{2\cdot (-1)^{3} - 9\cdot (-1) \cdot(-1) + 27\cdot 2}{54} = \dfrac{-2 - 9 + 54}{54} =$

$=\dfrac{43}{54} > 0$

$S = Q^{3} - R^{2} = \left( \dfrac{4}{9} \right)^{3} - \left( \dfrac{43}{54} \right)^{2} = \dfrac{64}{729} - \dfrac{1849}{2916} = \dfrac{256 - 1849}{2916} = -\dfrac{1593}{2916}=-\dfrac{59}{108}$

$\displaystyle \left \{ {{S < 0} \atop {Q > 0}} \right.$

$\phi = \dfrac{1}{3} Arch \left( \dfrac{|R|}{\sqrt{Q^{3}} } \right) = \dfrac{1}{3} Arch \left( \dfrac{\dfrac{43}{54} }{\sqrt{\dfrac{64}{729} } } \right) = \dfrac{1}{3} Arch \left( \dfrac{\dfrac{43}{54} }{\dfrac{8}{27} } \right) = \dfrac{1}{3} Arch \left( \dfrac{43 \cdot 27 }{54 \cdot 8 } \right) =$ $= \dfrac{1}{3} Arch \left( \dfrac{43}{16} \right) = ch\left( \dfrac{1}{3} Arch( 2,6875) \right)$

$x = -2R \sqrt{Q} \cdot ch(\phi) - \dfrac{a}{3} = -2\cdot \dfrac{43}{54} \cdot \sqrt{\dfrac{4}{9} } \cdot ch\left( \dfrac{1}{3} Arch( 2,6875) \right) - \dfrac{-1}{3} =$

$= -\dfrac{43}{27} \cdot \dfrac{2}{3}ch\left( \dfrac{1}{3} Arch( 2,6875) \right) + \dfrac{1}{3} = -\dfrac{86}{81} ch\left( \dfrac{1}{3} Arch( 2,6875) \right) + \dfrac{1}{3} \approx -1,21$

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