<span>Множество натуральных чисел, делителей числа 30: 1, 2, 3, 5, 6, 10, 15, 30.
П</span><span>одмножество простых чисел: 2, 3, 5.
(простые числа - числа, которые имеют только 2 делителя: оно делится на само себя и 1)</span>
2/(х-2)(х+2)+1/2(х-2)+7/2х(х+2)=0
ОДЗ:x<>2,-2(<> означает не равно)
приводим к общему знаменателю:
2*2х+х(х+2)+7(х-2)=0
4х+х^2+2х+7х-14=0
13х+х^2-14=0
х1=-14
х2=1
Ответ: -14; 1.
4. Под знак дифференциала постепенно загоняем: сначала косинус, затем двойку и наконец единицу, т.е.
![\frac{1}{2}d(2sinx+1) =cosxdx](https://tex.z-dn.net/?f=+%5Cfrac%7B1%7D%7B2%7Dd%282sinx%2B1%29+%3Dcosxdx)
, чтобы получился табличный интеграл от степенной функции.
![\int\limits^{ \pi /2}_0 { \sqrt{2sinx+1} * cosx} \, dx =\int\limits^{ \pi /2}_0 { \sqrt{2sinx+1} } \, d(sinx) = \\ \\ =\int\limits^{ \pi /2}_0 { \frac{1}{2} \sqrt{2sinx+1} } \, d(2sinx+1) = \frac{1}{2}\int\limits^{ \pi /2}_0 {(2sinx+1)^{ \frac{1}{2} } } \, d(2sinx+1) = \\ \\ = \frac{1}{2} \frac{2}{3} (2sinx+1)^{ \frac{3}{2}}= \frac{1}{3} (2sinx+1)^{ \frac{3}{2}}|_{0}^{\pi /2}= \\ \\ = \frac{1}{3} (2sin \frac{ \pi }{2} +1)^{ \frac{3}{2}} -\frac{1}{3} (2sin 0 +1)^{ \frac{3}{2}} =](https://tex.z-dn.net/?f=+%5Cint%5Climits%5E%7B+%5Cpi+%2F2%7D_0+%7B+%5Csqrt%7B2sinx%2B1%7D+%2A+cosx%7D+%5C%2C+dx+%3D%5Cint%5Climits%5E%7B+%5Cpi+%2F2%7D_0+%7B+%5Csqrt%7B2sinx%2B1%7D+%7D+%5C%2C+d%28sinx%29+%3D+%5C%5C++%5C%5C+%3D%5Cint%5Climits%5E%7B+%5Cpi+%2F2%7D_0+%7B+%5Cfrac%7B1%7D%7B2%7D+%5Csqrt%7B2sinx%2B1%7D+%7D+%5C%2C+d%282sinx%2B1%29+%3D+%5Cfrac%7B1%7D%7B2%7D%5Cint%5Climits%5E%7B+%5Cpi+%2F2%7D_0+%7B%282sinx%2B1%29%5E%7B+%5Cfrac%7B1%7D%7B2%7D+%7D+%7D+%5C%2C+d%282sinx%2B1%29+%3D+%5C%5C++%5C%5C+%3D+%5Cfrac%7B1%7D%7B2%7D++%5Cfrac%7B2%7D%7B3%7D+%282sinx%2B1%29%5E%7B+%5Cfrac%7B3%7D%7B2%7D%7D%3D+%5Cfrac%7B1%7D%7B3%7D+%282sinx%2B1%29%5E%7B+%5Cfrac%7B3%7D%7B2%7D%7D%7C_%7B0%7D%5E%7B%5Cpi+%2F2%7D%3D+%5C%5C++%5C%5C+%3D+%5Cfrac%7B1%7D%7B3%7D+%282sin+%5Cfrac%7B+%5Cpi+%7D%7B2%7D+%2B1%29%5E%7B+%5Cfrac%7B3%7D%7B2%7D%7D+-%5Cfrac%7B1%7D%7B3%7D+%282sin+0+%2B1%29%5E%7B+%5Cfrac%7B3%7D%7B2%7D%7D+%3D)
![= \frac{1}{3} (2sin \frac{ \pi }{2} +1)^{ \frac{3}{2}} -\frac{1}{3} (2sin 0 +1)^{ \frac{3}{2}} = \frac{1}{3} \sqrt{27} -\frac{1}{3}= \sqrt{3} -\frac{1}{3}](https://tex.z-dn.net/?f=%3D+%5Cfrac%7B1%7D%7B3%7D+%282sin+%5Cfrac%7B+%5Cpi+%7D%7B2%7D+%2B1%29%5E%7B+%5Cfrac%7B3%7D%7B2%7D%7D+-%5Cfrac%7B1%7D%7B3%7D+%282sin+0+%2B1%29%5E%7B+%5Cfrac%7B3%7D%7B2%7D%7D+%3D+%5Cfrac%7B1%7D%7B3%7D+%5Csqrt%7B27%7D+-%5Cfrac%7B1%7D%7B3%7D%3D+%5Csqrt%7B3%7D++-%5Cfrac%7B1%7D%7B3%7D)
5.
![\int\limits^4_2 { \frac{1}{x-1} } \, dx =\int\limits^4_2 { \frac{1}{x-1} } \, d(x-1) =ln(x-1)|_{2}^{4}=ln3-ln1=ln3](https://tex.z-dn.net/?f=+%5Cint%5Climits%5E4_2+%7B+%5Cfrac%7B1%7D%7Bx-1%7D+%7D+%5C%2C+dx+%3D%5Cint%5Climits%5E4_2+%7B+%5Cfrac%7B1%7D%7Bx-1%7D+%7D+%5C%2C+d%28x-1%29+%3Dln%28x-1%29%7C_%7B2%7D%5E%7B4%7D%3Dln3-ln1%3Dln3)
Arccos(x)`=-1/(√(1-x²)
arccos²(x)`=-2*arccos(x)/√(1-x²)
1/arccos²(x)=(1`*arccos²(x)-1*arccos²(x)`)/arccos⁴(x)=
=(0-(-2*arccos(x)/√(1-x²))/arccos⁴(x)=2*arccos(x)/(arccos⁴(x)*√(1-x²)).