Неопределённость ∞/∞ раскрываем делением числителя и знаменателя на эн в максимальной степени, т.е. на
![n^4](https://tex.z-dn.net/?f=n%5E4)
![\lim_{n \to \infty} \frac{n^4+5n^2-1}{10n^3-3n+2}= \lim_{n \to \infty} \frac{1+ \frac{5}{n^2}- \frac{1}{n^4} }{ \frac{10}{n} - \frac{3}{n^3} + \frac{2}{n^4} }=\frac{1+ \frac{5}{oo^2}- \frac{1}{oo^4} }{ \frac{10}{oo} - \frac{3}{oo^3} + \frac{2}{oo^4} }= \\ \\ =\frac{1+ 0-0}{0-0+0}= \frac{1}{0} =oo](https://tex.z-dn.net/?f=+%5Clim_%7Bn+%5Cto+%5Cinfty%7D++%5Cfrac%7Bn%5E4%2B5n%5E2-1%7D%7B10n%5E3-3n%2B2%7D%3D+%5Clim_%7Bn+%5Cto+%5Cinfty%7D++%5Cfrac%7B1%2B+%5Cfrac%7B5%7D%7Bn%5E2%7D-+%5Cfrac%7B1%7D%7Bn%5E4%7D+%7D%7B+%5Cfrac%7B10%7D%7Bn%7D+-+%5Cfrac%7B3%7D%7Bn%5E3%7D+%2B+%5Cfrac%7B2%7D%7Bn%5E4%7D+%7D%3D%5Cfrac%7B1%2B+%5Cfrac%7B5%7D%7Boo%5E2%7D-+%5Cfrac%7B1%7D%7Boo%5E4%7D+%7D%7B+%5Cfrac%7B10%7D%7Boo%7D+-+%5Cfrac%7B3%7D%7Boo%5E3%7D+%2B+%5Cfrac%7B2%7D%7Boo%5E4%7D+%7D%3D+%5C%5C++%5C%5C+%3D%5Cfrac%7B1%2B+0-0%7D%7B0-0%2B0%7D%3D+%5Cfrac%7B1%7D%7B0%7D+%3Doo)
(бесконечность)
A) x(x+2) = (x^2 + 2x + 1) - 1 = (x+1)^2 - 1.
Получаем:
(x+1)^2 > (x+1)^2 - 1 - Доказано
б)
<span>a^2+1 >= 2(3a-4)
</span>a^2+1 >= 6a - 8
a^2 - 6a + 9 >= 0
(a-3)^2 >= 0 - ДОКАЗАНО
2/x-<span>5 +14/x=3
(2+14)/x-5=3
16/x-5=3
16:x=8
x=2</span>
(x^3)^2-9x^3+8=0, x^3=a. a^2-9a+8=0, D=81-4*1*8=49, a1=(9-7)/2, a2=(9+7)/2. a1=1,a2=8. x^3=1, x1=1. x^3=8, x2=2. Ответ: x1+x2=1+2=3.
2+x≤5x-8 /+8
2+8+x≤5x-8+8
10+x≤5x / -x
10+x-x≤5x-x
10≤4x
4x≥10 /:4
x≥10:4
x≥5:2
x≥2,5
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