Если , то получим линейное неравенство: Полученный промежуток не включает в себя заданыый . Рассматриваем случай, когда - имеем квадратное неравенство. Заданное неравенство ">0", в зависимости от знака старшего коэффициента общие решения неравенства можно записать в виде: - если старший коэффициент больше 0: - если старший коэффициент меньше 0: Вывод: необходимо рассмотреть случай с положительным старшим коэффициентом: , тогда Решаем неравенство. Приравниваем левую часть к нулю: Получившийся дискриминант всегда больше 0, т.к. Чтобы получившийся ответ включал интервал х>3, необходимо потребовать выполнение следующего условия: Так как в рассматриваемом случае , то можно перейти к следующему неравенству: Итоговое решение с учетом рассматриваемого ограничения : Искомое минимальное целое значение ответ: 2
1 yesterday i had breakfast at the canteen. 2 last summer we went to the black sea. 3 he didn't call me last week. 4 why did this girl come to our office yesterday? 5 when did you buy tickets last time? 6 last june the boys played football every morning. 7 on sunday my parents were at the theatre. 8 i didn't take part at the conference last month. 9 he lived in kiev in his childhood. 10 where did he send his message? 2 трансформируйте предложения в past simplе 1 he often played chess. 2 our classes began at 9 a. m. 3 she was always busy. 4 what did you do in the evening? 5 my sister didn’t study at school, she worked at the hospital. 6 my friend prefered tea to coffee. 7 i found many stories in this magazine. 8 he was in my room, wasn't he? 9 why did you always ask so many questions? 10 jasmine seldom visited her grandparents. 3 образуйте отрицательную форму предложений 1 in summer she didn't get up early. 2 mr. white didn't write a lot of letters yesterday. 3 on friday they weren't at the station. 4 we didn't enjoy our time in rome. 5 they didn't watch a football match last weekend. 6 the girl didn't open a box at once. 7 i didn't leave my keys on the table. 8 the engineer didn't go on business trip on monday. 9 my friend didn't want to move to another flat. 10 i didn't find a surprise in the present-box. 4 поставьте вопросы к предложениям. 1 he learned english at school. (where? what language? why? or …? ) where did he learn english? what language did he learn? why did he learn english? did he learn english or french? 2 the train was five hours late. (what? why? ) what was 5 hours late? why was the train five hours late? 3 i was going to stay at my aunt’s in the crimea. (where? who? or …? ) where was i going to stay in the crimea? who was going to stay at the aunt's in the crimea? was i going to stay at my aunt's or in the hotel in the crimea? 4 they stayed in moscow only for two days. (why? where? how long? ) why did they stay in m only for 2 days? where did they stay? how long did they stay in m? 5 i bought a very beautiful dress in this shop last weekend. (where? what? when? ) where did i buy a very beautiful dress last weekend? what did i buy in this shop last weekend? when did i buy this beautiful dress in the shop? 6 they traveled to cyprus by sea. (how? where? or …? ) how did they travel to c? where did they travel to? did they travel to c or ibiza? 7 i didn’t find your telephone number. (what? why? ) what didn't i find? why didn't i find your telephone number? 8 in his childhood he was always a leader among his friends. (what? when? who? ) what was he in his childhood? when was he a leader among his friends? who was always a leader among his friends in his childhood? 9 he caught a cold last month. (what? when? why? ) what did he catch? when did he catch a cold? why did he catch a cold? 10 we did our lessons together. (what? how? ) what did we do together? how did we do our lessons?
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, то получим линейное неравенство:
.
- имеем квадратное неравенство. 
, тогда 
![4-3a+\sqrt{a^2-a+1} \leq 0 \\\ \sqrt{a^2-a+1} \leq 3a-4 \\\ \begin{cases} a^2-a+1 \leq (3a-4)^2 \\ 3a-4\ \textgreater \ 0 \right \end{cases} \\\ \begin{cases} a^2-a+1 \leq 9a^2-24a+16 \\ 3a\ \textgreater \ 4 \right \end{cases} \\\ \begin{cases} 8a^2-23a+15 \geq 0 \\ a\ \textgreater \ \frac{4}{3} \right \end{cases} \\\ \begin{cases} a\in(-\infty;1]\cup[ \frac{15}{8} ;+\infty) \\ a\ \textgreater \ \frac{4}{3} \right \end{cases}](/tpl/images/0507/6430/22606.png)
