Question

a)[(3,4+1,9-6,7):0,1+3,9]•4+40
b){[(2²la puterea n+8 la n+16 la n):4 la n-1:4 la n-1]:2-1}:2 la n-1;n apartine lui IN
ex 2) determinaţi numerele x y z ştiind că x y z sunt proportionale cu 2,4 respectiv 6 si x+1/x+z=5/24

Answer 1

A) [(3,4+1,9-6,7):0,1+3,9]•4+40 =
=[(-1,4):0,1+3,9)]•4+40 =
=(-14+3,9)•4+40 =
=(-10,1)•4+40 =
=-40,4+40=0,4

b){[(2^2n+8^n+16^n):4^(n-1):4^(n-1)]:2-1}:2^(n-1) =
={[2^2n+2^3n+2^4n):2^(4n-4)]:2-1}:2^(n-1) =
=[(2^(4-2n)+2^(4-n)+2^4)]:2-1}:2^(n-1) =
=[2^(3-2n)+2^(3-n)+2^3-2⁰]:2^(n-1) =
=2^(4-3n)+2^(4-2n)+2^(4-n)-2^(1-n)


x=2k
y=4k
z=6k
(2k+1)/(2k+6k)=5/24
(2k+1)/8k=5/24
48k+24=40k
8k=-24
k=-3
x=2k=2*(-3)=-6
y=4k=4*(-3)=-12
z=6k=6*(-3)=-18
verificare
(-6+1)/(-6-12)=-5/(-24)=5/24