Question

x²+5x-14≥0пожалуйста решите

Answer 1

Ответ:

Объяснение:

$x^{2} +5x-14\geq 0\\x^{2} -2x+7x-14\geq 0\\x(x-2)+7(x-2)\geq 0\\(x+7)(x-2)\geq 0\\x+7\geq 0\\x\geq -7\\x-2\geq 0\\x\geq 2$

+ + + (-7) - - - (2) + + +

x ∈ (-∞; -7] ∪ [2; +∞)

Answer 2

$\displaystyle x^2+5x-14\geq 0\\x^2+7x-2x-14\geq 0\\x(x+7)-2(x+7)\geq 0\\(x+7)(x-2)\geq 0\\\\\left \{ {{x+7\geq 0} \atop {x-2\geq 0}} \right. \rightarrow \left \{ {{x\geq -7} \atop {x\geq 2}} \right. \rightarrow x \in [2,+ \infty)\\ \\\left \{ {{x+7\leq 0} \atop {x-2\leq 0}} \right. \rightarrow \left \{ {{x\leq -7} \atop {x\leq 2}} \right. \rightarrow x\in (- \infty,-7]\\\\x \in (- \infty,-7] \cup[2,+ \infty)$