Question

Упростите : (0,2 m^2 * n^3)^3 * 100 m^4 * n^7

Найдите сумму кубов данного уравнения: x^2 + 3x + 1 = 0

Answer 1

(0,2 m^2 * n^3)^3 * 100 m^4 * n^7=0,008m^{6}*n^{9}*100m^{4}*n^{7}=

=0,8m^{10}*n^{16}

 

x^2 + 3x + 1 = 0

D=3^{2}-4*1*1=5

x1=(-3+sqrt{5})/2

x2=(-3-sqrt{5})/2

 

$x_{1}^{3}+x_{2}^{3}=(frac{-3+sqrt{5}}{2} )^{3}+(frac{-3-sqrt{5}}{2} )^{3}=$$(frac{-3+sqrt{5}-3-sqrt{5}}{2})((frac{-3+sqrt{5}}{2})^{2}-(frac{-3+sqrt{5}}{2})(frac{-3-sqrt{5}}{2})+(frac{-3-sqrt{5}}{2})^{2})=$$=-3*frac{9+5-6sqrt{5}-9+5+9+5+6sqrt{5}}{4}=frac{-3*24}{4}=-24$