Question

已知：实数a.b.c满足|a+1|+(3b+1)^2+|c-2|=0,使求：（3ab）^2乘以（-a^2c）^3乘以（6

Answer 1

因为|a+1|≥0,(3b+1)^2≥0,|c-2|≥0,而|a+1|+(3b+1)^2+|c-2|=0,故|a+1|=0,(3b+1)^2=0,|c-2|=0,即a=-1,b=-（1/3）,c=2,所以（3ab）^2*（-a^2c）^3*（6ab^2）^2=(3*-1*-(1/3))^2*(1^(2*2))*(6*-1*-(1/3)^2)^2=1*1(-2/3)^2=4/9