Задача 32530 Решить самостоятельную работу для 7...
Условие
Решение
2 б)
3
(4y-1)(4y+1)=(4y)^2-1^2=16y^2-1;
4
(x+0,3)(0,3-x)=(0,3+x)(0,3-x)=(0,3)^2-x^2=0,09-x^2
5
(-2a+b^2)(2a+b^2)=(b^2-2a)(b^2+2a)=(b^2)^2-(2a)^2=b^4-4a^2
6
16m^2-(1/49)n^4=(4m)^2-(n^2/7)^2=(4m-(n^2/7))(4m+(n^2/7))
7
(m+2)^2-(m-3)(m+3)=(m^2+4m+4)-(m^2-3^2)=
=m^2+4m+4-m^2+9=4m+13
8
654,68^2-345,32^2=(654,68-345,32)(654,68+354,32)=309,36*1=309,36
9
((1/2)m^2-(1/5)n^2)((1/2)m^2+(1/5)n^2)-((1/2)m^2-(1/5)n^2)^2=
=((1/2)m^2)^2 -((1/5)n^2)^2-[b]([/b]((1/2)m^2)^2-2*(1/2)m^2*(1/5)n^2+((1/5)n^2)^2[b])[/b]=
=(1/4)m^4-(1/25)n^4-(1/4)m^4+(1/5)m^2n^2-(1/25)n^4=
=(1/5)m^2n^2- (2/25)n^4
10
(a-2b)(a+2b)=a^2-(2b)^2=a^2-4b^2
(a^2-4b^2)(a^2+4b^2)=(a^2)^2-(4b^2)^2=a^4-16b^4
(a^4-16b^4)(a^4+16b^4)=(a^4)^2-(16b^4)^2=a^(8)-256b^(8)
О т в е т. a^(8)-256b^(8)