Question

Cate solutii naturale are ecuatia : x(x+1)+x(x+3)+x(x+5)+x(x+7)+x(x+2019)=1+2+3+...+2018

Answer 1

x∈ IN ?!

x(x+1)+x(x+3)+x(x+5)+x(x+7)+x(x+2019)=1+2+3+...+2018

x· [ (x+1)+ (x+3)+ (x+5)+(x+7)+... + (x+2019)]=1+2+3+...+2018

x· ( x+ 1+x+ 3+x+ 5+ x+ 7)+... +x+ 2019) =1+2+3+...+2018

x·[ ( x+x+ ... +x)+ ( 1+ 3+ ... + 2019)] =1+2+3+...+2018

x·[ ( x+x+ ... +x)+ ( 1+ 3+ ... + 1 009·2+1)] = 2 018·( 2018+1):2

x·[ ( x· 1 009)+ (1 009+1)² ] = 2 018· 2 019:2

x·(1 009·x+  1 010²)= 1 009· 2 019

1 009x² + 1 010²x =  1 009· 2 019

1 009x² + 1 010²x - 1 009· 2 019 = 0

D = b²- 4ac 

D = ( 1 010²)²- 4( 1 009)·( -1 009·2 019)

D= 1 010⁴ + 4· 1009²·2 019

D ≠ pp, deci x₁, x₂ ∉ IN

D > 0 , soluţii 2 ∈ IR

Answer 2

ecuatia estede grad2 ;deci poate avea 0,1 sau2  solutii

fie x=n o solutie a acestei ecuatii
atunci n(n+1+n+3+n+5+...n+2019)=2018*2019/2
avem de la 2*0+1 pan la 2*1009+1 =1010 numere impare si 1010 de n
si 1010/2 sume egale cu 1+2019=3+2017=...=2020

n(1010n+ 2020*1010/2)=1009*2019
n(1010n+1010²)=1009*2019
n*1010 (n+1010)=1009*2019
dac n e natural, numarul din stanga se termina in 0,de la factorul  1010,  numarul din dreapta s termina in 1(de la 9*9, ultimele cifre ale lui 1009 si 2019)
deci NU EXISTA solutie naturala