Ответ:
1
[tex]E_{1} =\frac{2x-5}{3} \\E_{2} =\frac{4x+1}{5}[/tex]
a)
[tex]\frac{2x-5}{3} < \frac{4x+1}{5} \\10x-25 < 12x+3\\-28 < 2x\\x > -14[/tex]
б)
[tex]\frac{2x-5}{3}+\frac{4x+1}{5} \geq 0\\\frac{10x-25+12x+3}{15}\geq 0\\ 22x-22\geq 0\\x\geq 1[/tex]
в)
[tex]\left \{ {{\frac{2x-5}{3} > 0} \atop {\frac{4x+1}{5} \leq 0}} \right. \\\left \{ {{2x-5 > 0} \atop {4x+1\leq 0}} \right. \\\left \{ {{x > 2.5} \atop {x\leq -0.25}} \right.[/tex]
решений нет
2)
[tex]a_{n} = 7n+2\\a_{n+1} = 7(n+1)+2\\a_{n+1}-a_{n} = 7(n+1)+2 - 7n -2 = 7n+7+2-7n-2 = 7 > 0[/tex]
значит последовательность возрастающая
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Answers & Comments
Ответ:
1
[tex]E_{1} =\frac{2x-5}{3} \\E_{2} =\frac{4x+1}{5}[/tex]
a)
[tex]\frac{2x-5}{3} < \frac{4x+1}{5} \\10x-25 < 12x+3\\-28 < 2x\\x > -14[/tex]
б)
[tex]\frac{2x-5}{3}+\frac{4x+1}{5} \geq 0\\\frac{10x-25+12x+3}{15}\geq 0\\ 22x-22\geq 0\\x\geq 1[/tex]
в)
[tex]\left \{ {{\frac{2x-5}{3} > 0} \atop {\frac{4x+1}{5} \leq 0}} \right. \\\left \{ {{2x-5 > 0} \atop {4x+1\leq 0}} \right. \\\left \{ {{x > 2.5} \atop {x\leq -0.25}} \right.[/tex]
решений нет
2)
[tex]a_{n} = 7n+2\\a_{n+1} = 7(n+1)+2\\a_{n+1}-a_{n} = 7(n+1)+2 - 7n -2 = 7n+7+2-7n-2 = 7 > 0[/tex]
значит последовательность возрастающая