Question

se dau numerele naturale nenule a, b, c , d care satisfac conditiile: b=2a, 3b=2c, 4c=3d.Aratati ca b la puterea 2 = ad. Aflati numerele a, b, c ,d stiind ca a la puterea 2 + b la puterea 2+c la puterea 2+ d la putetea 2= 480

Answer 1

4c = 3d => c = 3/4*d
3b = 2c => b = 2/3 * c = 2/3 * 3/4 * d = 2/4 * d = 1/2 * d
b = 2a => b^2 = b*b = 2a * 1/2*d = a*d

a^2 + b^2 + c^2 + d^2 = 480
Dar: 4c = 3d, de unde d = 4c/3
3b = 2c, de unde b = 2c/3
b = 2a, sau 2c/3 = 2a, de unde a = 2c/6 = c/3

Introducem d = 4c/3, b = 2c/3 si a = c/3 in prima relatie:
(c/3)^2 + (2c/3)^2 + c^2 + (4c/3)^2 = 480
c^2/9 + 4c^2/9 + c^2 + 16c^2/9 = 480
21c^2/9 + 9c^2/9 = 480
30c^2/9 = 480
10c^2/3 = 480
10c^2 = 1440
c^2 = 144
c = plus sau minus 12

c = 12
a = c/3 = 4
b = 2c/3 = 8
d = 4c/3 = 16

c = -12
a = c/3 = -4
b = 2c/3 = -8
d = 4c/3 = -16