1.
![a_{n} =3n-5 \\ a_{5} =3*5-5=10 \\ a_{25} =3*25-5=70](https://tex.z-dn.net/?f=a_%7Bn%7D+%3D3n-5+%5C%5C+%0Aa_%7B5%7D+%3D3%2A5-5%3D10+%5C%5C+%0Aa_%7B25%7D+%3D3%2A25-5%3D70)
2.
![a_{n} = a_{1} +d(n-1) \\ a_{6} = a_{1} +5d \\ a_{10} = a_{1} +9d \\ \left \{ {{a_{1} +5d=1} \atop {a_{1} +9d=13}} \right. \\ (2)-(1): 4d=12 \\ d=3 \\ a_{1}=1-5*3=-14 \\ S_{n} = \frac{2a_{1}+ d(n-1)}{2}*n \\ S_{20} = \frac{2*(-14)+3(20-1)}{2}*20=10(-28+57)=290](https://tex.z-dn.net/?f=a_%7Bn%7D+%3D+a_%7B1%7D+%2Bd%28n-1%29+%5C%5C+%0Aa_%7B6%7D+%3D+a_%7B1%7D+%2B5d+%5C%5C+%0Aa_%7B10%7D+%3D+a_%7B1%7D+%2B9d+%5C%5C+%0A+%5Cleft+%5C%7B+%7B%7Ba_%7B1%7D+%2B5d%3D1%7D+%5Catop+%7Ba_%7B1%7D+%2B9d%3D13%7D%7D+%5Cright.++%5C%5C+%0A%282%29-%281%29%3A+4d%3D12+%5C%5C+d%3D3+%5C%5C+%0Aa_%7B1%7D%3D1-5%2A3%3D-14+%5C%5C+S_%7Bn%7D+%3D+%5Cfrac%7B2a_%7B1%7D%2B+d%28n-1%29%7D%7B2%7D%2An+%5C%5C+S_%7B20%7D+%3D+%5Cfrac%7B2%2A%28-14%29%2B3%2820-1%29%7D%7B2%7D%2A20%3D10%28-28%2B57%29%3D290)
3. <span>а3=7 и а5=1.
</span>
![a_{n} = a_{1} +d(n-1) \\ \left \{ {{ a_{1}+2d=7} \atop {a_{1}+4d=1}} \right.](https://tex.z-dn.net/?f=a_%7Bn%7D+%3D+a_%7B1%7D+%2Bd%28n-1%29+%5C%5C++%5Cleft+%5C%7B+%7B%7B+a_%7B1%7D%2B2d%3D7%7D+%5Catop+%7Ba_%7B1%7D%2B4d%3D1%7D%7D+%5Cright.+)
<span>вычитаем из второго уравнения первое
2d=-6
d=-3
</span>
![a_{1}+2*(-3)=7 \\ a_{1}=13](https://tex.z-dn.net/?f=+a_%7B1%7D%2B2%2A%28-3%29%3D7+%5C%5C+++a_%7B1%7D%3D13)
![a_{17} = a_{1}+16d=13-16*3=-35](https://tex.z-dn.net/?f=a_%7B17%7D+%3D+a_%7B1%7D%2B16d%3D13-16%2A3%3D-35+)
<span>
</span>4.
![a_{n} =3n+2 \\ d=3 \\ a_{1} =3*1+2=5 \\ a_{30}=3*30+2=92 \\ S_{n} = \frac{a_{1}+ a_{n} }{2}*n \\ S_{30} = \frac{5+92}{2}*30=15*97=1455](https://tex.z-dn.net/?f=a_%7Bn%7D+%3D3n%2B2+%5C%5C+d%3D3+%5C%5C+a_%7B1%7D+%3D3%2A1%2B2%3D5+%5C%5C+a_%7B30%7D%3D3%2A30%2B2%3D92+%5C%5C+S_%7Bn%7D+%3D+%5Cfrac%7Ba_%7B1%7D%2B+a_%7Bn%7D+%7D%7B2%7D%2An+%5C%5C+S_%7B30%7D+%3D+%5Cfrac%7B5%2B92%7D%7B2%7D%2A30%3D15%2A97%3D1455)
4/5=0,80
0,75<0,80
а 11/7 и 13/9 я не помню. извини
<span>2, 4, 8, 16, 32, 64, 128 и т.д. </span>
<span>значит последняя цифра 2^50 будет 4</span>
![y = \sqrt{\dfrac{1}{2}x^{2} -3x} - \dfrac{1}{2x}](https://tex.z-dn.net/?f=y%20%3D%20%5Csqrt%7B%5Cdfrac%7B1%7D%7B2%7Dx%5E%7B2%7D%20-3x%7D%20-%20%5Cdfrac%7B1%7D%7B2x%7D)
Данная функция может существовать, если выполнится два условия (ОДЗ):
![\left \{ {\bigg{\dfrac{1}{2}x^{2} - 3x \geqslant 0} \atop \bigg{2x\neq 0 \ \ \ \ \ \ \ \ }} \right.](https://tex.z-dn.net/?f=%5Cleft%20%5C%7B%20%7B%5Cbigg%7B%5Cdfrac%7B1%7D%7B2%7Dx%5E%7B2%7D%20-%203x%20%5Cgeqslant%200%7D%20%5Catop%20%5Cbigg%7B2x%5Cneq%200%20%5C%20%5C%20%5C%20%5C%20%5C%20%5C%20%5C%20%5C%20%7D%7D%20%5Cright.)
Решим по отдельности каждое условие:
![1) \ 2x \neq 0; \ x\neq 0](https://tex.z-dn.net/?f=1%29%20%5C%202x%20%5Cneq%200%3B%20%5C%20x%5Cneq%200)
![2) \ \dfrac{1}{2}x^{2} - 3x \geqslant 0\\\dfrac{1}{2}x^{2} - 3x = 0 \ \ \ \ \ \ | \cdot 2\\x^{2} - 6x = 0\\x(x - 6) = 0\\x = 0; \ \ \ \ \ x = 6\\x \in (-\infty; \ 0] \cup [6; \ +\infty)](https://tex.z-dn.net/?f=2%29%20%5C%20%5Cdfrac%7B1%7D%7B2%7Dx%5E%7B2%7D%20-%203x%20%5Cgeqslant%200%5C%5C%5Cdfrac%7B1%7D%7B2%7Dx%5E%7B2%7D%20-%203x%20%3D%200%20%5C%20%5C%20%5C%20%5C%20%5C%20%5C%20%7C%20%5Ccdot%202%5C%5Cx%5E%7B2%7D%20-%206x%20%3D%200%5C%5Cx%28x%20-%206%29%20%3D%200%5C%5Cx%20%3D%200%3B%20%5C%20%5C%20%5C%20%5C%20%5C%20x%20%3D%206%5C%5Cx%20%5Cin%20%28-%5Cinfty%3B%20%5C%200%5D%20%5Ccup%20%5B6%3B%20%5C%20%2B%5Cinfty%29)
Объединим эти два условия и получим:
![x \in (-\infty; \ 0) \cup [6; \ +\infty)](https://tex.z-dn.net/?f=x%20%5Cin%20%28-%5Cinfty%3B%20%5C%200%29%20%5Ccup%20%5B6%3B%20%5C%20%2B%5Cinfty%29)
Ответ: ![D(y): \ x \in (-\infty; \ 0) \cup [6; \ +\infty)](https://tex.z-dn.net/?f=D%28y%29%3A%20%5C%20x%20%5Cin%20%28-%5Cinfty%3B%20%5C%200%29%20%5Ccup%20%5B6%3B%20%5C%20%2B%5Cinfty%29)
Всего с 10 до 99 90 двузначных чисел
из них дают остаток 1
8n+1 n=2 8*2+1=17
n=3; 8*3+1=25...
17;25;33;41;49;57;65;73;81;89;97 всего 11 чисел
Р=11/90 по-моему так!