Question

Vă salut,

Solicit sprijin pentru exercițiul de mai jos:

Determinati numerele x, y, z, știind că 1/x=2/y=3/z=4/t și x+y/2+z/3+t/4=1/8.

P.S. După nenumărate încercări și potriveli, eu am obținut: x=8, y=16, z=24, t=32.

Answer 1

Răspuns:

1/x =2/y=3/z=4/t =k coeficient de proportionalitate

x=1/k

y=2/k

z=3/k

t=4/k

x+y/2+z/3+t/4=1/8

1/k+(2/k)/2+(3/k)/3+(4/k)/4=1/8

1/k+1/k+1/k+1/k=1/8

4/k = 1/8

k=4×8/1=>k=32

x=1/k=>x=1/32

y=2/k=2/32 =>y=1/16

z=3/k=>z=3/32

t=4/k>t=4/32=>t=1/8

verificare:

1/32+1/16:2/1+3/32:3/1+1/8:4/1=

1/32+1/16×1/2+3/32×1/3+1/8×1/4=

1/32+1/32+1/32+1/32=(4/32)⁽⁴=1/8

Answer 2

Explicație pas cu pas:

$\frac{1}{x} = \frac{2}{y} = \frac{3}{z} = \frac{4}{t}$

$y = 2x \\ z = 3x \\ t = 4x$

$x + \frac{y}{2} + \frac{z}{3} + \frac{t}{4} = \frac{1}{8} \\ x + \frac{2x}{2} + \frac{3x}{3} + \frac{4x}{4} = \frac{1}{8}\\ x + x + x + x = \frac{1}{8} \\ = > 4x = \frac{1}{8} = > x = \frac{1}{32}$

$y = 2 \times \frac{1}{32} = \frac{1}{16} \\ z = 3 \times \frac{1}{32} = \frac{3}{32} \\ t = 4 \times \frac{1}{32} = \frac{1}{8}$

verificare:

$x + \frac{y}{2} + \frac{z}{3} + \frac{t}{4} = \frac{1}{32} + \frac{ \frac{1}{16} }{2} + \frac{ \frac{3}{32} }{3} + \frac{ \frac{1}{8} }{4} = \frac{1}{32} + \frac{1}{32} + \frac{1}{32} + \frac{1}{32} = \frac{4}{32} = \frac{1}{8}$