Question

№1

упрастите выражения:

а) b*(b^3)^4/b^9

б) 9x^2y^3-x^2y^3-10x^2y^3

в) (3x^2y)^4*(3xy^3)^2

г) (c^4)^5*c^8/(c^7)^4

№2

Вычислите:

 21^12/(7^4)^3* (3^2_^4

№3

Сравните значения выражений :

(3/2)^6* (2/3)^5 и 125^0

Answer 1

$b*(b^3)^4/b^9=b^{13}/b^9=b^4\9x^2y^3-x^2y^3-10x^2y^3=-2x2y^3\(3x^2y)^4 * (3xy^3)^2=(3^4x^8y^4)(3^2x^2y^6)=3^6x^{10}y^{10}=729x^{10}y^{10}\(c^4)^5 * c^8 / (c^7)^4=c^{20}*c^8 / c^{28}=c^{28}/c^{28}=1$$b*(b^3)^4/b^9=b^{13}/b^9=b^4\9x^2y^3-x^2y^3-10x^2y^3=-2x2y^3\(3x^2y)^4 * (3xy^3)^2=(3^4x^8y^4)(3^2x^2y^6)=3^6x^{10}y^{10}=729x^{10}y^{10}\(c^4)^5 * c^8 / (c^7)^4=c^{20}*c^8 / c^{28}=c^{28}/c^{28}=1$

 

 

$frac{21^{12}}{(7^4)^3 * (3^2)^4}=frac{3^{12}*7^{12}}{7^{12} * 3^8}=3^4=81$

 

 

$(3/2)^6*(2/3)^5;125^0\(3/2)^6*(3/2)^{-5};1\(3/2)^{6-5};1\3/2;1\1,5>1\(3/2)^6*(2/3)^5>125^0$