Помогите решить! 1)(16а³б²-4а²б³+а²б²):(4а²б²) 2)(10а⁴+15а³):5а²-18а³:(-6а) при а=-3/2

[latex] \frac{16a^3b^2-4a^2b^3+a^2b^2}{4a^2b^2} = \frac{4a^2b^2(4a-b+1)}{4a^2b^2}=4a-b+1 [/latex] ____________________________________________________________________________ [latex] \frac{10a^4+15a^3}{5a^2}- \frac{18a^3}{-6a} = \frac{5a^2(2a^2+3a)}{5a^2}+3a^2 [/latex] [latex]=5a^2+3a[/latex] при a=-3/2? тогда [latex]5*(- \frac{3}{2})^2+3*(- \frac{3}{2}) = \frac{5*6}{4}- \frac{9}{2}= \frac{30-18}{4}= \frac{12}{4}=3 [/latex]