Question

1).(1/8)^2 / 2^5*(1/2)^10=
2).2x*(x^2-8x-2)=
3).(-3y^2+0,6y)*(-1,5y^3)=
4).5x+3(x-1)=6x+11

Answer 1

Объяснение:

1)

$\frac{(\frac{1}{8})^2}{2^5*(\frac{1}{2})^{10} } =\frac{((\frac{1}{2})^3)^2 }{2^5*2^{-10}} =\frac{(\frac{1}{2})^6 }{2^{-5}} =\frac{2^{-6}}{2^{-5}} =2^{-1}=\frac{1}{2} =0,5.$

2)

$2x*(x^2-8x-2)=2x^3-16x^2-4x.$

3)

$(-3y^2+0,6y)*(-1,5y^3)=4,5y^5-0,9y^4.$

4)

$5x+3*(x-1)=6x+11\\5x+3x-3=6x+11\\8x-3=6x+11\\2x=14|:2\\x=7.$