Задачи с урока c ответами
- \((x-1)(x-3)<0\)
- \(\displaystyle\frac{(x+4)(3-x)}{2(x+1)}\ge0\)
- \(\displaystyle\frac{-4(x-1)^2(3-x)^3}{(x+1)^4}\ge0\)
- \(\displaystyle\frac{2}{x-1}\ge3\)
- \(\displaystyle\frac{x+7}{x+2}>x-1\)
- \(2x^2-3x+1\le0\)
- \(\displaystyle\frac{(x^2+x+1)^3(x+5)^5}{x^2+1}>0\)
- \(\displaystyle\frac{x^3-1}{(x^4-16)(1-x)}\le0\)
- \(\displaystyle\frac{1}{x+1}-\frac{2}{x^2-x+1}\le\frac{1-2x}{x^3+1}\)
- \(\displaystyle\frac{(x-3)^2(x-7)^3(x+1)}{(x-2)(x+4)^4}\ge0\)
- \(x^2+1>\displaystyle\frac{x^2-5}{x^2+2}\)
- \(\displaystyle\frac{25}{x^2-4x}>x^2-4x\)
Домашнее задание с ответами
- \((3-x)(x-5)\le0\)
- \(2x^2+3x+37<(x+7)^2\)
- \((x^3-1)(x^4-16)<0\)
- \(\displaystyle\frac{(x-1)^2(x-2)^3(x+3)}{(x-2)(x+7)^4}\ge0\)
- \(\displaystyle\frac{x+1}{3x-5}\le\frac{1}{3}\)
- \(\displaystyle\frac{1}{x}+x\le1\)
- \((\displaystyle\frac{1}{x^2}-4)(\frac{1}{x}+2)\le0\)
- \(\displaystyle\frac{7x-16}{x^2-7x+12}\ge-1\)
- Найдите наименьшее положительное решение неравенства \(\displaystyle\frac{x^2-5x+4}{(x^2+2)(x+2)}\le0\)
- Найдите наибольшее целое решение неравенства \(\displaystyle\frac{x}{x-1}-\frac{2}{x+1}-\frac{8}{x^2-1}<0\)
- \(x^2(13-x^2)\ge36\)
- —
- \((-\infty;3]\cup [4;+\infty)\)
- (-1; 12)