Ирационално уравнение

[Гост]

Да се реши уравнението:

[tex]\sqrt[3]{ (x-1)^{2 } } + \sqrt[3]{ (2 + x)^{2 } } = \sqrt[3]{- x^{2 } - x +2 } + 3[/tex]

[ammornil]

$\sqrt[3]{(x-1)^{2 } } + \sqrt[3]{(2 + x)^{2 }} = \sqrt[3]{- x^{2 } - x +2 } + 3 \\[6pt] \text{ДМ}:\quad \forall{x}\in{\mathbb{R}} \\[12pt] \sqsupset{} \begin{cases} \sqrt[3]{(x-1)}= u \Rightarrow x= u^{3} +1 \\ \sqrt[3]{(x+2)}= v \Rightarrow x= v^{3} -2 \end{cases} \\[6pt] \sqrt[3]{(x-1)^{2 } } + \sqrt[3]{(2 + x)^{2 }} = \sqrt[3]{- x^{2 } - x +2 } + 3 \Rightarrow u^{2} +v^{2}= -u\cdot{v} +3 \Leftrightarrow u^{2} +u\cdot{v} +v^{3} =3 \\[6pt] v^{3} -u^{3}= (x+2) -(x-1)= 3 \\[6pt] v^{3} -u^{3} =(v-u)\underbrace{(v^{2} +u\cdot{v} +u^{3})}_{=3}= 3 |\div{3} \Rightarrow v-u= 1 \Rightarrow v= u+1 \\[12pt] v^{3} -u^{3}= 3 \Leftrightarrow (u+1)^{3} -u^{3}= 3 \Leftrightarrow u^{3} +3u^{2} +3u +1 -u^{3}= 3 \\[6pt] \Leftrightarrow 3u^{2} +3u -2= 0 \Rightarrow u_{1,2}= \dfrac{-3\pm{}\sqrt{3^{2}-4\cdot{3}\cdot{(-2)}}}{2\cdot{3}} =\dfrac{-3\pm{}\sqrt{33}}{6} \\[6pt] u_{1,2}^{3}= \dfrac{(-3)^{3}\pm{}3\cdot{(-3)^{2}}\cdot{\sqrt{33}}+3\cdot{(-3)}\cdot{(\pm\sqrt{33})^{2}}\pm{(\sqrt{33})^3}}{216} =\dfrac{-27\pm{27\sqrt{33}-297\pm{33\sqrt{33}}}}{216} \\[6pt] u_{1,2}^{3}= \dfrac{-324\pm{60}\sqrt{33}}{216}= -\dfrac{3}{2}\pm\dfrac{5\sqrt{33}}{18} \Rightarrow x_{1,2}= u_{1,2}^{3}+1 \Rightarrow $ $$ x_{1}= -\dfrac{1}{2} -\dfrac{5\sqrt{33}}{18} \\[6pt] x_{2}= -\dfrac{1}{2} +\dfrac{5\sqrt{33}}{18} $$

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$- x^{2 } - x +2= -(x^{2} +x -2) \\[6pt] \hspace{4em} x^{2} +x -2 \rightarrow x_{1,2}=\dfrac{-1\pm\sqrt{1^{2}-4\cdot{1}\cdot{(-2)}}}{2\cdot{1}}= \dfrac{-1\pm{3}}{2} \\[12pt] -(x^{2} +x -2)= -1\cdot{}(x-1)\cdot{}(x+2) \Rightarrow \sqrt[3]{- x^{2 } - x +2}=-\sqrt[3]{x-1}\cdot{}\sqrt[3]{x+2} $