Question

Для продвинутых!!!!!!!!!!!!!!!!!!!!!

Answer 1

Ответ:

(см. объяснение)

Объяснение:

$4(x^2+2)^4+81(x^3+1)^2\le36(x^7+4x^5+x^4+4x^3+4x^2+4)\\4(x^2+2)^4+81(x^3+1)^2\le36(x^3(x^4+4x^2+4)+(x^4+4x^2+4))\\4(x^2+2)^4+81(x^3+1)^2\le36(x^4+4x^2+4)(x^3+1)\\4(x^2+2)^4+81(x^3+1)^2\le36(x^2+2)^2(x^3+1)\\4(x^2+2)^4-36(x^2+2)^2(x^3+1)+81(x^3+1)^2\le0\\(2(x^2+2)^2-9(x^3+1))^2\le0\\2(x^2+2)^2-9(x^3+1)=0\\2(x^4+4x^2+4)-9x^3-9=0\\2x^4-9x^3+8x^2-1=0\\2x^4-2x^3-7x^3+7x^2+x^2-x+x-1=0\\2x^3(x-1)-7x^2(x-1)+x(x-1)+(x-1)=0\\(x-1)(2x^3-7x^2+x+1)=0\\(x-1)(2x^3-x^2-6x^2+3x-2x+1)=0$

$(x-1)(x^2(2x-1)-3x(2x-1)-(2x-1))=0\\(x-1)(2x-1)(x^2-3x-1)=0\\\left[\begin{array}{c}x-1=0\\2x-1=0\\x^2-3x-1=0\end{array}\right,\\\\\\1)\\x-1=0\\x=1\\\\2)\\2x-1=0\\2x=1\\x=\dfrac{1}{2}\\\\3)\\x^2-3x-1=0\\D=9+4=13\\\sqrt{D}=\sqrt{13}\\x_{1,2}=\dfrac{3\pm\sqrt{13}}{2}\\\\\left[\begin{array}{c}x=1\\x=\dfrac{1}{2}\\x=\dfrac{3+\sqrt{13}}{2}\\x=\dfrac{3-\sqrt{13}}{2}\end{array}\right;$

Неравенство решено!