решение задания смотри на фотографии
Применим формулы понижения степени:
![\sin^2\alpha=\dfrac{1-\cos 2\alpha}{2};~~~~~ \cos^2\alpha=\dfrac{1+\cos2\alpha}{2}](https://tex.z-dn.net/?f=%5Csin%5E2%5Calpha%3D%5Cdfrac%7B1-%5Ccos+2%5Calpha%7D%7B2%7D%3B~~~~~+%5Ccos%5E2%5Calpha%3D%5Cdfrac%7B1%2B%5Ccos2%5Calpha%7D%7B2%7D)
![1)~ \sin^215=\dfrac{1-\cos(2\cdot15)}{2}=\dfrac{1-\cos30}{2};\\ \\ 2)~ \cos^2\bigg(\dfrac{\pi}{4}-\alpha\bigg)=\dfrac{1+\cos\left(2\cdot \left(\frac{\pi}{4}-\alpha\right)\right)}{2}=\dfrac{1+\cos\left(\frac{\pi}{2}-2\alpha\right)}{2}](https://tex.z-dn.net/?f=1%29~+%5Csin%5E215%3D%5Cdfrac%7B1-%5Ccos%282%5Ccdot15%29%7D%7B2%7D%3D%5Cdfrac%7B1-%5Ccos30%7D%7B2%7D%3B%5C%5C+%5C%5C+2%29~+%5Ccos%5E2%5Cbigg%28%5Cdfrac%7B%5Cpi%7D%7B4%7D-%5Calpha%5Cbigg%29%3D%5Cdfrac%7B1%2B%5Ccos%5Cleft%282%5Ccdot+%5Cleft%28%5Cfrac%7B%5Cpi%7D%7B4%7D-%5Calpha%5Cright%29%5Cright%29%7D%7B2%7D%3D%5Cdfrac%7B1%2B%5Ccos%5Cleft%28%5Cfrac%7B%5Cpi%7D%7B2%7D-2%5Calpha%5Cright%29%7D%7B2%7D)
Решение во вложениииииииииииииииииииииииииииии
Ответ:
Объяснение:
1)
(1/3)²=1/9
OTBET: Г
2)
B^8 : b^4=b^(8-4)=b^4
OTBET: Б
3)
(3²*3^5*3^6) / 3^11 =3^(2+5+6)/3^11 = 3^13/3^11=3^(13-11)=3²
4)
(6a³b²)^4 *(8a^2b)^5 / 81*(2a^7b^4)³ = =(6^4a^12b^8*8^5a^10b^5)/(3^4*2^3a²1b^12=
=(2^4*3^4*a^22b^13(2³)^5 ) /2³*3^4a^21b^12= (2^19*3^4a^22b^13)/ 2³*3^4a^21b^12=
=2^16 ab
1)а)=х(у+2)+5(У+2)=(х+5)(у+2)
б)=3а^2-3аz-3az+3z^2=3a(a-z)-3z(a-z)=(3a-3z)(a-z)=3(a-z)(a-z)=3(a-z)^2