Объяснение:
(-2a3)^2*b^2*(3c^2)^2*a^3=
=4a^6*b^2*9c^4*a^3=36a^9b^2c^4
36*(2)^9*(1/4)^2*(1/2)^4=
=36*512*1/16*1/16=72
1)
![cos \alpha = \frac{12}{13}](https://tex.z-dn.net/?f=cos+%5Calpha+%3D+%5Cfrac%7B12%7D%7B13%7D+)
α -угол первой четверти.
В первой четверти все тригонометрические функции имеют знак "+".
![sin \alpha = \sqrt{1-cos^2 \alpha }= \sqrt{1-( \frac{12}{13} )^2}=](https://tex.z-dn.net/?f=sin+%5Calpha+%3D+%5Csqrt%7B1-cos%5E2+%5Calpha+%7D%3D+%5Csqrt%7B1-%28+%5Cfrac%7B12%7D%7B13%7D+%29%5E2%7D%3D++)
![= \sqrt{1- \frac{144}{169} }= \sqrt{ \frac{25}{169} }= \frac{5}{13}](https://tex.z-dn.net/?f=%3D+%5Csqrt%7B1-+%5Cfrac%7B144%7D%7B169%7D+%7D%3D+%5Csqrt%7B+%5Cfrac%7B25%7D%7B169%7D+%7D%3D+%5Cfrac%7B5%7D%7B13%7D+++)
![tg \alpha = \frac{sin \alpha }{cos \alpha }= \frac{ \frac{5}{13} }{ \frac{12}{13} }= \frac{5}{12}](https://tex.z-dn.net/?f=tg+%5Calpha+%3D+%5Cfrac%7Bsin+%5Calpha+%7D%7Bcos+%5Calpha+%7D%3D+%5Cfrac%7B+%5Cfrac%7B5%7D%7B13%7D+%7D%7B+%5Cfrac%7B12%7D%7B13%7D+%7D%3D+%5Cfrac%7B5%7D%7B12%7D+++)
2)
![tg( \frac{ \pi }{4}+ \alpha )= \frac{tg \frac{ \pi }{4}+tg \alpha }{1-tg \frac{ \pi }{4}tg \alpha }= \frac{1+tg \alpha }{1-tg \alpha }=](https://tex.z-dn.net/?f=tg%28+%5Cfrac%7B+%5Cpi+%7D%7B4%7D%2B+%5Calpha++%29%3D+%5Cfrac%7Btg+%5Cfrac%7B+%5Cpi+%7D%7B4%7D%2Btg+%5Calpha++%7D%7B1-tg+%5Cfrac%7B+%5Cpi+%7D%7B4%7Dtg+%5Calpha++%7D%3D+%5Cfrac%7B1%2Btg+%5Calpha+%7D%7B1-tg+%5Calpha+%7D%3D++)
![= \frac{1+ \frac{5}{12} }{1- \frac{5}{12} }= \frac{ \frac{17}{12} }{ \frac{7}{12} }= \frac{17}{7}](https://tex.z-dn.net/?f=%3D+%5Cfrac%7B1%2B+%5Cfrac%7B5%7D%7B12%7D+%7D%7B1-+%5Cfrac%7B5%7D%7B12%7D+%7D%3D+%5Cfrac%7B+%5Cfrac%7B17%7D%7B12%7D+%7D%7B+%5Cfrac%7B7%7D%7B12%7D+%7D%3D+%5Cfrac%7B17%7D%7B7%7D+++)
Ответ: 17/7.
sin(a+b)-cosa *sinb sina*cosb+cosa*sinb-cosa*sinb sina*cosb
<span>a = 5
b = 12
c = 8
S = 2(ab + bc + ac) = 2(5</span>×12+12×8+5×8) = 2(60+96+40) = 392
Log^2_{√6)216=log^2_{6^1/2}216=2*log^2_{6}216=2*9=18.