Question

Срочно помогите пожалуйста,два упражнения,даю что могу,15 баллов

Answer 1

Объяснение:

3.

$a)\ 16x-8x^2=8x*(2-x).\\b)\ -10ab-25a=-5a*(2b+5).\\c)\ 23bd-23d=23d*(b-1).\\d)\ 121am-77m=11m*(11a-7).\\e)\ 40pt-72t^2=8t*(5p-9t).\\f)\ -54bd+63b^2d^2=9bd*(7bd-6).\\g)\ a^5+5a^3=a^3*(a^2+5).\\h)\ 3m^4+24m^2=3m^2*(m^2+8).$

4.

$a)\ -24b^7-44b^4=-6b^4*(4b^3+11).\\b)\ 96b^8-72b^9=24b^8*(4-3b).\\c)\ 8x+12y-16z=4*(2x+3y-4z).\\d)\ 18a-27b-63c=9*(2a-3b-7c).\\e)\ 108a+72b-36c=18*(6a+4b-2c).\\f)\ 84m-91n-49k=7*(12m-13n-7k).$

Answer 2

$16x - {8x}^{2} = 8x(2 - x) \\ - 10ab - 25a = - 5a(2b + 5) \\ 23bd - 23d = 23d(b - 1) \\ 121am + 77m = 11m(11a + 7) \\ 40pt - {72t}^{2} = 8t(5p - 9t) \\ - 54bd + 63 {b}^{2} {d}^{2} = - 9bd( 6 - 7bd) \\ {a}^{5} + {5a}^{3} = {a}^{3}( {a}^{2} + 5) \\ 3 {m}^{4} + {24m}^{2} = {3m}^{2} ( {m}^{2} + 8)$

$- 24 {b}^{7} - 44 {b}^{4} = - 4 {b}^{4} (6 {b}^{3} + 11) \\ 96 {b}^{8} - 72 {b}^{9} = 24 {b}^{8} (4 - 3b) \\ 8x + 12y - 16z = 4(2x + 3y - 4z) \\ 18a - 27b - 63c = 9(2a - 3b - 7c) \\ 108a + 72b - 36c = 36(3a + 2b - c) \\ 84m - 91n - 49k = 7(12m - 13n - 7k)$