12.已知正数aaa,bbb满足:a2−2a+2+1=a+2b+4b2+1\sqrt{{a}^{2}-2a+2}+1=a+2b+\sqrt{4{b}^{2}+1}a2−2a+2+1=a+2b+4b2+1,则以下结论中 (1)a+2b=1a+2b=1a+2b=1 (2)a+2b=2a+2b=2a+2b=2 (3)1a+2b\frac{1}{a}+\frac{2}{b}a1+b2的最小值为9 (4)1a+2b\frac{1}{a}+\frac{2}{b}a1+b2的最小值为3 正确结论个数为((( )))