Предмет: Математика, автор: wwwarkhipovru

Неравенства с модулями

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Ответы

Автор ответа: NNNLLL54

$1)\; \; |3x+1| \leq 2-5x\; \; \; \Leftrightarrow \; \; \; -(2-5x) \leq 3x+1\leq 2-5x\\\\5x-2-1 \leq 3x \leq 2-5x-1\; ,\; \; \; 5x-3 \leq 3x \leq 1-5x\; \; \Leftrightarrow \\\\\left \{ {{5x-3\leq 3x} \atop {3x\leq 1-5x}} \right. \; \; \left \{ {{2x \leq 3} \atop {8x \leq 1}} \right. \; \; \left \{ {{x \leq \frac{3}{2}} \atop {x \leq \frac{1}{8}}} \right. \; \; \Rightarrow \; \; x\leq \frac{1}{8}\; ,\; \; x\in (-\infty ,\frac{1}{8}\, ]\\\\2)\; \; |-x-5| \leq x^2+1\; \; \; \Leftrightarrow \; \; \; -(x^2+1)\leq -x-5\leq x^2+1$

$\left \{ {{-x^2-1 \leq -x-5} \atop {-x-5 \leq x^2+1}} \right. \; \; \left \{ {{x^2-x-4 \geq 0} \atop {x^2+x+4 \geq 0}} \right. \; \; \left \{ {{(x-\frac{1-\sqrt{17}}{2})(x-\frac{1+\sqrt{17}}{2})\geq 0} \atop {x\in (-\infty ,+\infty )}} \right. \; \; \Rightarrow \\\\x\in (-\infty ,\frac{1-\sqrt{17}}{2})\cup (\frac{1+\sqrt{17}}{2},+\infty )\\\\P.S.\; \; x^2-x-4=0\; ,\; \; x_{1,2}=\frac{1\pm \sqrt{17}}{2}\\\\x^2+x+4=0\; ,\; \; D=1-16=-15\ \textless \ 0\; \; \; net\; \; kornej$

$3)\; \; |x^2-4|\ \textgreater \ 2x\; \; \; \Rightarrow \; \; \; \left \{ {{x^2-4\ \textgreater \ 2x} \atop {x^2-4\ \textless \ -2x}} \right.\; \; \left \{ {{x^2-2x-4\ \textgreater \ 0} \atop {x^2+2x-4\ \textless \ 0}} \right. \\\\\left \{ {{(x-1)^2-5\ \textgreater \ 0} \atop {(x+1)^2-5\ \textless \ 0}} \right. \; \; \left \{ {{(x-1-\sqrt5)(x-1+\sqrt5)\ \textgreater \ 0} \atop {(x+1-\sqrt5)(x+1+\sqrt5)\ \textless \ 0}} \right. \; \; \left \{ {{x\in (-\infty ;\, 1-\sqrt5)\cup (1+\sqrt5;\, +\infty )} \atop {x\in (-1-\sqrt5;-1+\sqrt5)}} \right. \Rightarrow \\\\(-1-\sqrt5)\approx -3,2\; \; ,\; \; (1+\sqrt5)\approx 3,2\\\\(-1-\sqrt5)\approx -3,2\; \; ,\; \; (-1+\sqrt5)\approx 1,2\\\\x\in (-1-\sqrt5\, ;\, 1-\sqrt5)$

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