Найдите корни уравнения
Извлечем кубический корень из обеих частей равенства (решения при этом не теряются)
Ответ: х=2
![\sqrt[3]{((x-3)^{2})^3}=\sqrt[3]{-(x^{2}-2x-1)^{3}} \\ (x-3)^{2}=-(x^{2} -2x-1);\\ x^{2} -6x+9+x^{2} -2x-1=0;\\ 2x^{2} -8x+8=0;|*(\frac{1}{2}) \\ x^{2} -4x+4=0;\\ (x-2)^{2}=0;\\ x=2\\](https://tex.z-dn.net/?f=+%5Csqrt%5B3%5D%7B%28%28x-3%29%5E%7B2%7D%29%5E3%7D%3D%5Csqrt%5B3%5D%7B-%28x%5E%7B2%7D-2x-1%29%5E%7B3%7D%7D+%5C%5C++++%28x-3%29%5E%7B2%7D%3D-%28x%5E%7B2%7D+-2x-1%29%3B%5C%5C++x%5E%7B2%7D+-6x%2B9%2Bx%5E%7B2%7D+-2x-1%3D0%3B%5C%5C++2x%5E%7B2%7D+-8x%2B8%3D0%3B%7C%2A%28%5Cfrac%7B1%7D%7B2%7D%29+%5C%5C+x%5E%7B2%7D++-4x%2B4%3D0%3B%5C%5C++%28x-2%29%5E%7B2%7D%3D0%3B%5C%5C+x%3D2%5C%5C+++ "\sqrt[3]{((x-3)^{2})^3}=\sqrt[3]{-(x^{2}-2x-1)^{3}} \\ (x-3)^{2}=-(x^{2} -2x-1);\\ x^{2} -6x+9+x^{2} -2x-1=0;\\ 2x^{2} -8x+8=0;|*(\frac{1}{2}) \\ x^{2} -4x+4=0;\\ (x-2)^{2}=0;\\ x=2\\")