|a-2|+(b+1/4)^2=0∵|a-2|≥0,(b+1/4)^2≥0∴a-2=0,b+1/4=0a=2,b=-1/4(a^b-2ab)-2(3ab^2+4ab)=a^b-6ab^2-10ab=2^(1/4)-6*2*(1/4)^2-10*2*1/4=2^(1/4)-3/4-5=2^(1/4)-23/4？？？？？？？？？？？？？？？？？？？？？？？？？？？？？？？？？？？題目是不是這樣：已知|a-2|+[(b+1)/4]^2=0，求(a^b-2ab)-2(3ab^2+4ab)的值|a-2|+(b+1/4)^2=0因為|a-2|≥0，[(b+1)/4]^2≥0所以a-2=0，(b+1)/4=0a=2，b=3(a^b-2ab)-2(3ab^2+4ab)=a^b-6ab^2-10ab=2^3-6*2*3^2-10*2*3=8-108-60=-160