Question

Помогите, пожалуйста, с неравенствами.....

Answer 1

$1)\; \; \sqrt[3]{x^3-3x^2+2x+8}\ \textless \ 1+x\\\\x^3-3x^2+2x+8\ \textless \ 1+3x+3x^2+x^3\\\\6x^2+x-7\ \textless \ 0\; ,\; \; D=169\; ,\; \; x_1=-\frac{7}{6}\; ,\; x_2=1\\\\+++(-\frac{7}{6})---(1)+++\\\\x\in (-\frac{7}{6};1)\\\\2)\; \; (x-3)^{11}\ \textgreater \ (x^2-4x+3)^{11}\\\\x-3\ \textgreater \ x^2-4x+3\\\\x^2-5x+6\ \textless \ 0\; ,\; \; x_1=2\; ,\; x_2=3\; \; (teor.\; Vieta)\\\\+++(2)---(3)+++\\\\x\in (2,3)$

$3)\; \; (\frac{3}{5})^{2-x}\ \textless \ (\frac{3}{5})^{3x-2}\\\\\frac{3}{5}\ \textless \ 1\; \; \Rightarrow \; \; 2-x\ \textgreater \ 3x-2\\\\4\ \textgreater \ 4x\; \; \Rightarrow \; \; 4x\ \textless \ 4\; ,\; x\ \textless \ 1\\\\x\in (-\infty ,1)\\\\4)\; \; 3^{cos^2x}\ \textgreater \ 3^{sin^2x+0,5}\\\\3\ \textgreater \ 1\; \; \Rightarrow \; \; cos^2x\ \textgreater \ sin^2x+0,5\\\\cos^2x\ \textgreater \ (1-cos^2x)+0,5}\\\\2cos^2x\ \textgreater \ 1,5\; \; \Rightarrow \; \; cos^2x\ \textgreater \ \frac{3}{4}\\\\cos^2x=\frac{1+cos2x}{2}\; \; \Rightarrow \; \; \frac{1+cos2x}{2}\ \textgreater \ \frac{3}{4}$

$1+cos2x\ \textgreater \ \frac{3}{2}\\\\cos2x\ \textgreater \ \frac{1}{2}\\\\-\frac{\pi}{3}+2\pi n\ \textless \ 2x\ \textless \ \frac{\pi}{3}+2\pi n\\\\-\frac{\pi}{6}+\pi n\ \textless \ x\ \textless \ \frac{\pi}{6}+\pi n\; ,\; \; n\in Z$