Question

№1 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$

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

$(3/2)^6*(2/3)^5;125^0\\(3/2)^6*(3/2)^{-5};1\\(3/2)^{6-5};1\\3/2;1\\1,51\\(3/2)^6*(2/3)^5125^0$