Explicatii:
La fiecare din subpunctele problemei A1) avem cate un interval inchis la stanga si la dreapta, care contine si la stanga si la dreapta cate o expresie E1 si E2, in functie de x.
Se cere sa gasim valorile lui x pentru care intervalele sunt multimi nevide.
Conditia pentru a fi multime nevida este:
E1 < E2
Nu putem accepta intervalul [5; 3]
Nici nu poate fi desenat pe axa Ox.
Putem accepta conditia
E1 ≤ E2
La egalitate vom avea de exemplu intervalul
[5; 5] care este egal cu {5} si care are un singur element.
Rezulta ca e o multime nevida.
Rezolvare:
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[tex]\displaystyle\bf\\a)~~I=[-2x+1;~1+3x]\\\\-2x+1\leq1+3x\\\\-2x+1-1-3x\leq0\\\\ -2x-3x\leq0\\\\-5x\leq0~~\Big|:(-5)~~(Se~schimba~sensul~inegalitatii)\\\\x\geq0\\\\x\in[0,~\infty)[/tex]
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[tex]\displaystyle\bf\\\\b)~~I=\left[\frac{x}{x+1},~\frac{1}{x} \right]\\\\\\\frac{x}{x+1}\leq\frac{1}{x}\\\\x^2\leq1(x+1)\\\\x^2\leq x+1\\\\x^2-x-1<0\\\\a=1>0 \implies~x^2-x-1~~este~negativa~intre~radacini.\\\\x_{12}=\frac{1\pm\sqrt{1+4}}{2}=\frac{1\pm\sqrt{5}}{2}\\\\x_1=\frac{1-\sqrt{5}}{2}\\\\x_2=\frac{1+\sqrt{5}}{2}\\\\x\in\left[\frac{1-\sqrt{5}}{2},~~\frac{1+\sqrt{5}}{2}\right]-\{0;~-1\}[/tex]
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[tex]\displaystyle\bf\\c)~~I=\left[\frac{x-1}{x+1},~\frac{2}{x+2}\right]\\\\\\\frac{x-1}{x+1}\leq\frac{2}{x+2}\\\\(x-1)(x+2)\leq2(x+1)\\\\x^2-x+2x-2\leq2x+2\\\\x^2-x+2x-2-2x-2\leq0\\\\x^2-x-4\leq0\\\\a=1>0 \implies~x^2-x-1~~este~negativa~intre~radacini.\\\\x_{12}=\frac{1\pm\sqrt{1+16}}{2}\\\\x_1=\frac{1-\sqrt{17}}{2}\\\\x_2=\frac{1+\sqrt{17}}{2}\\\\\\x\in\left[\frac{1-\sqrt{17}}{2},~~\frac{1+\sqrt{17}}{2}\right]-\{-1;~-2\}[/tex]
