Question

помогите 2 И 4
срочно​

Answer 1

2) (9c^2 - 4d^2) / (18c^2d - 12cd^2) = [(3c-2d)(3c+2d)] / [6cd(3c-2d)] = (3c+2d)/(6cd)

Подставляем c = 1 2/3 = 5/3; d = 0,5 = 1/2

(3c+2d)/(6cd) = (3*5/3+2*1/2)/(6*5/3*1/2) = (5+1)/5 = 6/5 = 1,2

4) (x^2+6xy+9y^2) / (4x^2+12xy) = (x+3y)^2 / [4x(x+3y)] = (x+3y)/(4x)

Подставляем x = - 0,2; y = - 0,6

(x+3y)/(4x) = (-0,2-3*0,6)/(-4*0,2) = (-0,2-1,8)/(-0,8) = 2/0,8 = 20/8 = 5/2 = 2,5