∵方程4x²-4(k+1)x+k²+1=0有两实根∴Δ=16(k+1)²-4*4*(k²+1)=32k+16-16=32k≥0∴k≥0x1+x2=k+1＞0x1*x2=(k²+1)/4＞0∴x1＞0,x2＞0∴|x1|+|x2|=x1+x2=k+1=2∴k=1