liczba naturalna to całkowita dodatnia n=1liczba n+1=2 liczba n+2=3 liczba n²+(n+1)²+(n+2)²=149 n²+n²+2n+1+n²+4n+4=149 3n²+6n+5-149=0 3n²+6n-144=0 Δ=b²-4ac=36+1728=1764 √Δ=42 n₁=(-b-Δ):2a=(-6-42):6=-8 odpada bo to nie liczba naturalna n₂=(-b+√Δ):2a=(-6+42):6=6 n=6 to liczby: n=6 n+1=6+1=7 n+2=6+2=8 spr. 6²+7²+8²=36+49+64=149 to liczby:6,7,8
Kolejne liczby naturalne to: n, N +1, n +2 zatem mamy n² +(n +1)² + ( n +2)² = 149 n² + n² + 2n +1 + n² + 4n + 4 = 149 3 n² + 6n + 5 = 149 3 n² + 6 n - 144 = 0 / : 3 n² + 2 n - 48 = 0 Δ = 4 - 4*1*(-48) = 4 + 192 = 196 √Δ = 14 n = [ -2 - 14]/ 2 = - 16/2 = - 8 < 0 - odpada n = [ -2 + 14]/2 = 12/2 = 6 Odp. n = 6 Te liczby to: 6, 7, 8. ===============================
n²+(n+1)²+(n+2)²=149 n²+n²+2n+1+n²+4n+4=149 3n²+6n-144=0 n²+2n-48=0 Δ=2²-4*1*(-48) Δ=4+192 Δ=196 √Δ=14 n₁=(-2-14)/(2*1) n₁=-16/2 n₁=-8 <-- odpada bo -8∉N n₂=(-2+14)/(2*1) n₂=12/2 n₂=6 To te liczby: n=6 n+1=7 n+2=8
