Алгебра, на картинке

Решите задачу:

$1)\; \; \frac{(3x^2+x-2)(x^2-x)}{(3x-2)(x^2-x+1)}\leq 0\\\\a)\; \; 3x^2+x-2=0\; ,\; D=25\; \; ,\; \; x_1=\frac{2}{3}\; ,\; \; x_2=-1\\\\3x^2+x-2=3(x-\frac{2}{3})(x+1)\\\\b)\; \; x^2-x+1=0\; ,\; \; D=-3\ \textless \ 0\; \; \to \; \; x^2-x+1\ \textgreater \ 0\\\\c)\; \; \frac{3(x-\frac{2}{3})(x+1)\cdot x\cdot (x-1)}{3\cdot (x-\frac{2}{3})} \leq 0\; ,\; \; x\ne \frac{2}{3}\\\\ (x+1)\cdot x\cdot (x-1) \leq 0\\\\---[-1\, ]+++[\, 0\, ]---[\, 1\, ]+++\\\\\underline {x\in (-\infty ,-1\, ]\cup [\, 0,\frac{2}{3})\cup (\frac{2}{3},1\, ]}$

$2)\; \; (x-1)(x^2+1)(x^3-1)(x^4+1)\ \textless \ 0\\\\a)\; \; x^2+1\ \textgreater \ 0\; ,\; \; \; x^4+1\ \textgreater \ 0\; \; \to \\\\b)\; \; (x-1)(x^3-1)\ \textless \ 0\\\\(x-1)(x-1)(x^2+x+1)\ \textless \ 0\\\\(x-1)^2(\underbrace {x^2+x+1}_{\ \textgreater \ 0})\ \textless \ 0\\\\x^2+x+1=0\; ,\; \; D=-3\ \textless \ 0\; \; \to \; \; x^2+x+1\ \textgreater \ 0\\\\c)\; \; (x-1)^2\ \textless \ 0\; \; \Rightarrow \; \; \underline {x\in \varnothing }\; ,\; \; t.k.\; \; (x-1)^2 \geq 0\; .$

Алгебра, ** картинке