[tex]\frac{1000}{x} +21 = (x+2)^{2}[/tex]
[tex]dla \ \ x = 40\\\\L = \frac{1000}{4}+21= 250 +21 = 271\\\\P = (4+1)^{2} = 5^{2} = 25\\\\L \neq P[/tex]
[tex]dla \ \ x = 10\\\\L = \frac{1000}{10}+21 = 100+21 = 121\\\\P = (10+1)^{2} = 11^{2} = 121\\\\\underline{L = P}[/tex]
[tex]dla \ \ x = 100\\\\L = \frac{1000}{100}+21 = 10+21 = 31\\\\P = (100+1)^{2} = 101^{2} = 10201\\\\L \neq P[/tex]
Odp. Rozwiązaniem tego równanianjest liczba 10.
