Y=x^3*cosx
y'=(x^3)'*cosx+x^3*(cosx)'=3x^2cosx-x^3sinx
по формуле синуса двойного угла
![\frac{10sin 16cos 16}{sin 32}=\\ \frac{5*2sin 16cos 16}{sin 32}=\\ \frac{5*sin (2*16)}{sin 32}=\\ \frac{5 sin 32}{sin 32}=5](https://tex.z-dn.net/?f=%5Cfrac%7B10sin+16cos+16%7D%7Bsin+32%7D%3D%5C%5C+%5Cfrac%7B5%2A2sin+16cos+16%7D%7Bsin+32%7D%3D%5C%5C+%5Cfrac%7B5%2Asin+%282%2A16%29%7D%7Bsin+32%7D%3D%5C%5C+%5Cfrac%7B5+sin+32%7D%7Bsin+32%7D%3D5)
А) = c²+3c-2c-6-c²=c-6
б) = 7x+8x+x²-64=x²+15x-64
в) = 4x²+20x-4x²-20x-25=-25
Кхм..
<span>Это случайно не 6 класс?
Лови,решение:
</span>х = 1, у = 1
<span>3x + ay = 5 </span>
<span>7x - by = 6
</span>
3 + а = 5 ⇒ а = 2
<span>7 - b = 6 ⇒ </span><span>b = 1
</span>Удачи!