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((( )))