rezolvati in R ecuatia x^2+(x+2)^2+(x+4)^2+...+(x+98)^2=(x+1)^2+(x+3)^2+...+(x+99)^2
Răspunsuri la întrebare
Fie A= x²+(x+2)²+...+(x+98)²
B=(x+1)²+(x+3)²+...(x+99)²
pt f(x) =x+a , crescatoare pr R
fie g(x)=x+a, f(x) :{0.1;2...99}
atunci f(0)<f(1)<f(2)<...<f(98)<f(99)
x+0<x+1<x+2<x+3<....x+98<x+99
si, prin ridicare la patrat
x²<(x+1)²
(x+1)²<(x+2)²
(x+2)²<(x+4)²
........................
(x+98)²<(x+99)²
A<B , ecuatia nu are solutii
pt x<-99
x<x+1<x+2<x+3<....<x+98<x+99<0
si x²>(x+1)²>0
(x+2)²>(x+3)²>0
.........................
(x+98)²>(x+99)²
A>B , ecuatia nu aresolutii
deci x∈(-99;0)
cum e ecuatia e de grad, va avea 0 , 1 sau 2 solutii
fie x=-49,5
( am ales la bunul simt media aritmerica a extremelor celor 2 siruri de la x=0 la x+99)
A=49,5²+47,5²+45,5²+..+3,5²+1,5²+0,5²+2,5²+4,5²....+46,5²+48,5²
B=48,5²+46,5²+....+2,5²+0,5²+1,5²+3,5²+...+47,5²+49,5²
x=-49,5 verifica
sorry, nu stiu sa demonstrez ca e si unica