Obliczmy
[tex] {x}^{2} + {8}^{2} = ( {x + 3} ) ^{2} \\ {x}^{2} + 64 = (x + 3)(x + 3) \\ {x}^{2} + 64 = {x}^{2} + 3x + 3x + 9 \\ - 6x = 9 - 64 \\ - 6x = - 55 \\ x = 9 \frac{1}{6} [/tex]
Zatem
[tex]x = 9 \frac{1}{6 } \\ x + 3 = 9 \frac{1}{6} + 3 = 12 \frac{1}{6} [/tex]
Myślę że pomogłem ;)
