Question

x,y,z dp 3,6,8 de unde relatia x+y+z=68​

Answer 1

Răspuns:

{x.y.z} d.p {3,6,8}

x+y+z=68

x×3=k=>k3

y×6=k=>k6

z×8=k=>8k

3k+6k+8k=68

17k=68=>k=4

x=3×4=12

y=6×4=24

z=8×4=32

Answer 2

Răspuns:

$x + y + z = 68 \: \: \: \: \: \: \frac{x}{3} = \frac{y}{6} = \frac{z}{8} \\ \frac{x}{3} + \frac{y}{6} + \frac{z}{8} = \frac{x + y + z}{3 + 6 + 8} = \frac{ {68}^{(17} }{ {17}^{(17} } = 4 \\ \frac{x}{3} = 4; \: x = 3 \times 4 = 12 \\ \frac{y}{6} = 4; \: y = 6 \times 4 = 24 \\ \frac{z}{8} = 4; \: z = 8 \times 4 = 32$

sper că te-am ajutat