Question

помогите пожалуйста.​

Answer 1

Формулы:

$(a+b)^2=a^2+2ab+b^2\\\\ (a-b)^2=a^2-2ab+b^2\\ (a^k)^m=a^{k+m}$

Решение:

$1) (a^2+2a)^2=(a^2)^2+2a^2*2a+(2a)^2 = a^4+4a^3+4a^2.\\ 2) (\frac{1}{4} m^{3} -12m)^{2} =(\frac{1}{4} m^{3} )^2-2*\frac{1}{4} m^{3}*12m+12m^2=\frac{1}{16} m^{6} -6m^4+12m^2.\\ 3) (1\frac{1}{3} p^{7} +3p^2)^{2}=(1\frac{1}{3} p^{7})^2+2*1\frac{1}{3} p^{7}*3p^2+(3p^2)^2 = \frac{16}{9} p^{14}+8p^9+9p^4.\\ 4) (7ab-2b^3)^2 = (7ab)^2-2*7ab*2b^3+(2b^3)^2=49a^2b^2-28ab^4+4b^6.\\ 5) (10p^6+\frac{1}{2}p^4a^3)^{2}=(10p^6)^2+2*10p^6*\frac{1}{2}p^4a^3+(\frac{1}{2}p^4a^3)^2=100p^{12}+10a^3p^{10}+\frac{1}{4} a^6p^8.$

$6) (0,2m^2n+15m^3n^4)^2 =(0,2m^2n)^2+2*0,2m^2n*15m^3n^4+(15m^3n^4)^2=(\frac{1}{5})^{2}*m^4n^2+6m^5n^5+225m^6n^8=\frac{1}{25}*m^4n^2+6m^5n^5+225m^6n^8.$