![\frac{30}{16x^2+6-24xy+9y^2}=\frac{30}{(4x-3y)^2+6}](https://tex.z-dn.net/?f=%5Cfrac%7B30%7D%7B16x%5E2%2B6-24xy%2B9y%5E2%7D%3D%5Cfrac%7B30%7D%7B%284x-3y%29%5E2%2B6%7D)
Дробь принимает наибольшее значение,когда знаменатель принимает своё наименьшее значение. Это будет тогда, когда первое слагаемое будет =0.
Наибольшее значение дроби равно 5.
2) ООФ: знаменатель дроби не=0.
х(х+4) не=0, ---> х не=0 , х не=-4
х Є (-беск,-4)U(-4,0)U(0,беск)
Добрый день! Решение см. фото.
((sin13a+sin15a)+(sin14a+sin16a))/((cos13a+cos15a)+(cos14a+cos16a))=
(2sin14a*cosa+2sin15a*cosa)/(2cos14acosa+2cos15acosa)=
=2cosa*(sin14a+sin15a)/2cosa(cos14a+cos15a)=
=(2sin29a/2*cosa/2)/(2cos29a/2*cosa/2)=tg(29a/2)
6а-b-2a-3b=4a-4b=4(a-b)
8x-3x+2x-5y=7x-5y
Ответ:
1006
Объяснение:
![-x^{2020}+x^{2019}-x^{2018}+...-x^{2}+x=\\ =x^{2019}(1-x)+x^{2017}(1-x)+x(1-x)=\\=x(1-x)(x^{2018}+x^{2016}+...+x^{2}+1)](https://tex.z-dn.net/?f=-x%5E%7B2020%7D%2Bx%5E%7B2019%7D-x%5E%7B2018%7D%2B...-x%5E%7B2%7D%2Bx%3D%5C%5C%20%3Dx%5E%7B2019%7D%281-x%29%2Bx%5E%7B2017%7D%281-x%29%2Bx%281-x%29%3D%5C%5C%3Dx%281-x%29%28x%5E%7B2018%7D%2Bx%5E%7B2016%7D%2B...%2Bx%5E%7B2%7D%2B1%29)
Если n есть чётное (n = 2k)
, при k>1
![x^{n}+1 = (x + 1)(x^{n-1} - x^{n-2} +...+ x - 1)](https://tex.z-dn.net/?f=x%5E%7Bn%7D%2B1%20%3D%20%28x%20%2B%201%29%28x%5E%7Bn-1%7D%20-%20x%5E%7Bn-2%7D%20%2B...%2B%20x%20-%201%29)
или же
![x^{n} = (x + 1)(x^{n-1} - x^{n-2} +...+ x - 1)-1](https://tex.z-dn.net/?f=x%5E%7Bn%7D%20%3D%20%28x%20%2B%201%29%28x%5E%7Bn-1%7D%20-%20x%5E%7Bn-2%7D%20%2B...%2B%20x%20-%201%29-1)
для n=2 обратная ситуация
![x^{2} = (x + 1)(x-1)+1](https://tex.z-dn.net/?f=x%5E%7B2%7D%20%3D%20%28x%20%2B%201%29%28x-1%29%2B1)
обозначим
![M_{n}=x^{n-1} - x^{n-2} +...+ x - 1](https://tex.z-dn.net/?f=M_%7Bn%7D%3Dx%5E%7Bn-1%7D%20-%20x%5E%7Bn-2%7D%20%2B...%2B%20x%20-%201)
в результате получим наш многочлен
![x(1-x)(x^{2018}+x^{2016}+...+x^{2}+1)=\\=-x(x-1)(((x+1)M_{2018}-1)+((x+1)M_{2016}-1)+...+((x+1)M_{4}-1)+((x+1)(x-1)+1)+1)=\\=-x(x-1)((x+1)M_{2018}+(x+1)M_{2016}+...+(x+1)M_{4}+(x+1)(x-1)+(-1)\frac{2018-2}{2} +1+1)=\\=x(x-1)(-(x+1)(M_{2018}+M_{2016}+...+M_{4}+(x-1))+1006)](https://tex.z-dn.net/?f=x%281-x%29%28x%5E%7B2018%7D%2Bx%5E%7B2016%7D%2B...%2Bx%5E%7B2%7D%2B1%29%3D%5C%5C%3D-x%28x-1%29%28%28%28x%2B1%29M_%7B2018%7D-1%29%2B%28%28x%2B1%29M_%7B2016%7D-1%29%2B...%2B%28%28x%2B1%29M_%7B4%7D-1%29%2B%28%28x%2B1%29%28x-1%29%2B1%29%2B1%29%3D%5C%5C%3D-x%28x-1%29%28%28x%2B1%29M_%7B2018%7D%2B%28x%2B1%29M_%7B2016%7D%2B...%2B%28x%2B1%29M_%7B4%7D%2B%28x%2B1%29%28x-1%29%2B%28-1%29%5Cfrac%7B2018-2%7D%7B2%7D%20%2B1%2B1%29%3D%5C%5C%3Dx%28x-1%29%28-%28x%2B1%29%28M_%7B2018%7D%2BM_%7B2016%7D%2B...%2BM_%7B4%7D%2B%28x-1%29%29%2B1006%29)