Question

решите методом интервалов:
$(x + 6)(x + 1) ^4 (x - 3) geqslant 0$


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Answer 1

$(x + 6)(x + 1) {}^{4} (x - 3) geqslant 0 \ x in ( - infty ; - 6] cup (1) cup [3; + infty )$