Odpowiedź:
[tex](\sqrt{2}x+1)^2=2(x^2+x+1)\\\\(\sqrt{2}x)^2+2\sqrt{2}x\cdot1+1^2=2x^2+2x+2\\\\2x^2+2\sqrt{2}x+1=2x^2+2x+2\\\\\not2x^2+2\sqrt{2}x-\not2x^2-2x=2-1\\\\2\sqrt{2}x-2x=1\\\\(2\sqrt{2}-2)x=1\ \ |:(2\sqrt{2}-2)\\\\x=\frac{1}{2\sqrt{2}-2}\\\\x=\frac{1}{2\sqrt{2}-2}\cdot\frac{2\sqrt{2}+2}{2\sqrt{2}+2}\\\\x=\frac{2\sqrt{2}+2}{(2\sqrt{2}-2)(2\sqrt{2}+2)}\\\\x=\frac{2\sqrt{2}+2}{(2\sqrt{2})^2-2^2}\\\\x=\frac{2\sqrt{2}+2}{4\cdot2-4}\\\\x=\frac{2\sqrt{2}+2}{8-4}\\\\x=\frac{\not2^1(\sqrt{2}+1)}{\not4_{2}}[/tex]
[tex]x=\frac{1}{2}(\sqrt{2}+1)\\\\\\Odp.B[/tex]
