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x+3=numero impar consecutivo
(x+1)²+(x+3)²=130
x²+2x+1+x²+6x+9=130
x²+x²+2x+6x+1+9=130
2x²+8x+10=130
2x²+8x+10-130=0
2x²+8x-120=0 dividimos todo entre 2
x²+4x-60=0
(x+10)(x-6)=0
x+10=0 x-6=0
x= -10 x=6
tomamos el valor positivo x=6
la solucion es:
x+1.......6+1=7
x+3.......6+3=9
los numeros son 7 y 9
................................
7²+9²=130
49+81=130
130=130