Representa em extens\u00e3o os seguintes conjuntos:<\/p>\n
- \n
- $A=\\left\\{ x\\in \\mathbb{Z}:3(x-1)>4(x+2)\\wedge -12\\le x+3 \\right\\}$<\/li>\n
- $B=\\left\\{ x\\in \\mathbb{N}:4x-9\\le x<2x+1 \\right\\}$<\/li>\n
- $C=\\left\\{ x\\in \\mathbb{R}:3<\\frac{x}{4}\\vee 2(x-3)<6x \\right\\}$<\/li>\n<\/ol>\nResolu\u00e7\u00e3o >><\/a><\/span><\/div><\/div>\n\n\nResolu\u00e7\u00e3o<\/b><\/span><\/p>\n
- \n
- $A=\\left\\{ x\\in \\mathbb{Z}:3(x-1)>4(x+2)\\wedge -12\\le x+3 \\right\\}$\n
Comecemos por resolver a condi\u00e7\u00e3o:
\n$$\\begin{array}{*{35}{l}}
\n\\begin{matrix}
\n3(x-1)>4(x+2) & \\wedge\u00a0 & -12\\le x+3\u00a0 \\\\
\n\\end{matrix} & \\Leftrightarrow\u00a0 & \\begin{matrix}
\n3x-3>4x+8 & \\wedge\u00a0 & -x\\le 15\u00a0 \\\\
\n\\end{matrix}\u00a0 \\\\
\n{} & \\Leftrightarrow\u00a0 & \\begin{matrix}
\n-x>11 & \\wedge\u00a0 & x\\ge -15\u00a0 \\\\
\n\\end{matrix}\u00a0 \\\\
\n{} & \\Leftrightarrow\u00a0 & \\begin{matrix}
\nx<-11 & \\wedge\u00a0 & x\\ge -15\u00a0 \\\\
\n\\end{matrix}\u00a0 \\\\
\n{} & \\Leftrightarrow\u00a0 & \\begin{matrix}
\nx\\ge -15 & \\wedge\u00a0 & x<-11\u00a0 \\\\
\n\\end{matrix}\u00a0 \\\\
\n\\end{array}$$
\nOra, o conjunto A \u00e9 constitu\u00eddo pelos n\u00fameros inteiros relativos<\/strong> pertencentes ao intervalo \\[\\left[ -15,+\\infty\u00a0 \\right[\\cap \\left] -\\infty ,-11 \\right[=\\left[ -15,-11 \\right[\\]
\n![\"\"](\"https:\/\/www.acasinhadamatematica.pt\/wp-content\/uploads\/2012\/01\/pag107-35a.gif\" "\\"Intervalos\\"")
<\/a>
\nLogo:
\n$$\\begin{array}{*{35}{l}}
\nA & = & \\left\\{ x\\in \\mathbb{Z}:3(x-1)>4(x+2)\\wedge -12\\le x+3 \\right\\}\u00a0 \\\\
\n{} & = & \\left\\{ x\\in \\mathbb{Z}:\\begin{matrix}
\nx\\ge -15 & \\wedge\u00a0 & x<-11\u00a0 \\\\
\n\\end{matrix} \\right\\}\u00a0 \\\\
\n{} & = & \\left\\{ -15,-14,-13,-12 \\right\\}\u00a0 \\\\
\n\\end{array}$$
\n\u00ad<\/p>\n<\/li>\n- $B=\\left\\{ x\\in \\mathbb{N}:4x-9\\le x<2x+1 \\right\\}$\n
Comecemos por resolver a condi\u00e7\u00e3o:
\n$$\\begin{array}{*{35}{l}}
\n4x-9\\le x<2x+1 & \\Leftrightarrow\u00a0 & \\begin{matrix}
\n4x-9\\le x & \\wedge\u00a0 & x<2x+1\u00a0 \\\\
\n\\end{matrix}\u00a0 \\\\
\n{} & \\Leftrightarrow\u00a0 & \\begin{matrix}
\n3x\\le 9 & \\wedge\u00a0 & -x<1\u00a0 \\\\
\n\\end{matrix}\u00a0 \\\\
\n{} & \\Leftrightarrow\u00a0 & \\begin{matrix}
\nx\\le 3 & \\wedge\u00a0 & x>-1\u00a0 \\\\
\n\\end{matrix}\u00a0 \\\\
\n{} & \\Leftrightarrow\u00a0 & \\begin{matrix}
\nx>-1 & \\wedge\u00a0 & x\\le 3\u00a0 \\\\
\n\\end{matrix}\u00a0 \\\\
\n\\end{array}$$
\nOra, o conjunto B \u00e9 constitu\u00eddo pelos n\u00fameros naturais<\/strong> pertencentes ao intervalo
\n\\[\\left] -1,+\\infty\u00a0 \\right[\\cap \\left] -\\infty ,3 \\right]=\\left] -1,3 \\right]\\]
\n![\"\"](\"https:\/\/www.acasinhadamatematica.pt\/wp-content\/uploads\/2012\/01\/pag107-35b.gif\" "\\"Intervalos\\"")
<\/a>
\nLogo:
\n$$\\begin{array}{*{35}{l}}
\nB & = & \\left\\{ x\\in \\mathbb{N}:4x-9\\le x<2x+1 \\right\\}\u00a0 \\\\
\n{} & = & \\left\\{ x\\in \\mathbb{N}:\\begin{matrix}
\nx>-1 & \\wedge\u00a0 & x\\le 3\u00a0 \\\\
\n\\end{matrix} \\right\\}\u00a0 \\\\
\n{} & = & \\left\\{ 1,2,3 \\right\\}\u00a0 \\\\
\n\\end{array}$$
\n\u00ad<\/p>\n<\/li>\n- $C=\\left\\{ x\\in \\mathbb{R}:3<\\frac{x}{4}\\vee 2(x-3)<6x \\right\\}$\n
Comecemos por resolver a condi\u00e7\u00e3o:
\n$$\\begin{array}{*{35}{l}}
\n\\begin{matrix}
\n\\underset{(4)}{\\mathop{3}}\\,<\\frac{x}{\\underset{(1)}{\\mathop{4}}\\,} & \\vee\u00a0 & 2(x-3)<6x\u00a0 \\\\
\n\\end{matrix} & \\Leftrightarrow\u00a0 & \\begin{matrix}
\n12<x & \\vee\u00a0 & 2x-6<6x\u00a0 \\\\
\n\\end{matrix}\u00a0 \\\\
\n{} & \\Leftrightarrow\u00a0 & \\begin{matrix}
\nx>12 & \\vee\u00a0 & -4x<6\u00a0 \\\\
\n\\end{matrix}\u00a0 \\\\
\n{} & \\Leftrightarrow\u00a0 & \\begin{matrix}
\nx>12 & \\vee\u00a0 & x>-\\frac{3}{2}\u00a0 \\\\
\n\\end{matrix}\u00a0 \\\\
\n{} & \\Leftrightarrow\u00a0 & \\begin{matrix}
\nx>-\\frac{3}{2} & \\vee\u00a0 & x>12\u00a0 \\\\
\n\\end{matrix}\u00a0 \\\\
\n\\end{array}$$
\nOra, o conjunto C \u00e9 constitu\u00eddo pelos n\u00fameros reais<\/strong> pertencentes ao intervalo
\n\\[\\left] -\\frac{3}{2},+\\infty\u00a0 \\right[\\cup \\left] 12,+\\infty\u00a0 \\right[=\\left] -\\frac{3}{2},+\\infty\u00a0 \\right[\\]
\n![\"\"](\"https:\/\/www.acasinhadamatematica.pt\/wp-content\/uploads\/2012\/01\/pag107-35c.gif\" "\\"Intervalos\\"")
<\/a>Nota<\/strong>: O intervalo representado a amarelo encontra-se sobreposto ao intervalo representado a azul.<\/p>\nLogo:
\n\\[\\begin{array}{*{35}{l}}
\nC & = & \\left\\{ x\\in \\mathbb{R}:3<\\frac{x}{4}\\vee 2(x-3)<6x \\right\\}\u00a0 \\\\
\n{} & = & \\left\\{ x\\in \\mathbb{R}:\\begin{matrix}
\nx>-\\frac{3}{2} & \\vee\u00a0 & x>12\u00a0 \\\\
\n\\end{matrix} \\right\\}\u00a0 \\\\
\n{} & = & \\left] -\\frac{3}{2},+\\infty\u00a0 \\right[\u00a0 \\\\
\n\\end{array}\\]<\/p>\n<\/li>\n<\/ol>\n - $B=\\left\\{ x\\in \\mathbb{N}:4x-9\\le x<2x+1 \\right\\}$\n
- $A=\\left\\{ x\\in \\mathbb{Z}:3(x-1)>4(x+2)\\wedge -12\\le x+3 \\right\\}$\n