Question

fie x,y,z invers proportionale cu 0,(3) ;0,25; 0,2
a) Verificati daca x²+y²=z²
b)Dem ca $\frac{2x+2z}{5y+z} =patratul unui numar rational$
c)Aflati numerele stiind ca[tex] \frac{2x+2z}{5y+z} =patratul unui numar rational
\frac{3}{x} + \frac{4}{y} + \frac{5}{z} =1
[/tex]
\frac{3}{x} + \frac{4}{y} + \frac{5}{z} =1
[/tex]

Answer 1

$\frac{3}{9} x= \frac{25}{100} y= \frac{2}{10} z=k$
$\frac{x}{3} = \frac{y}{4} = \frac{z}{5} =k$
x=3k
y=4k
z=5k
a) x²+y²=z²
9k²+16k²=25k²
25k²=25k² (A) 
b) $\frac{2*2k+2*5k}{5*4k+5k} = \frac{6k+10k}{20k+5k} = \frac{16k}{25k} = \frac{16}{25} - nr rational$
c) $\frac{3}{3k} + \frac{4}{4k} + \frac{5}{5k} =1$
$\frac{1}{k} + \frac{1}{k} + \frac{1}{k} =1$
$\frac{3}{k} =1$
k=3
x=3*3=9
y=3*4=12
z=3*5=15

Answer 2

a)
0,(3)=3/9=1/3
0,25=25/100=1/4
0,2=2/10=1/5
x/[1/(1/3)] =k   x=3k
y/[1/(1/4)] =k   y=4k
z/[1/(1/5)] =k   z=5k
x²+y²=(3k)²(4k)²=9k²+16k²=25k²
z²=(5k)²=25k²
rezulta ca x²+y²=z²

b)
(2x+2z)/(5y+z)=(2×3k+2×5k)/(5×4k+5k)=16k/25k=16/25=(4/5)²

c)
3/x+4/y+5/z=1
3/3k+4/4k+5/5k=1
1/k+1/k+1/k=1
3/k=1
k=3
x=3k=3×3=9
y=4k=4×3=12
z=5k=5×3=15