作业帮计算[（2又3/5）^0]+[2^（-2）]*[(2又1/4)^-1/2]-[(0.01)^0.5]
来源：学生作业帮助网 编辑：作业帮 时间：2024/05/06 02:17:13
![计算[（2又3/5）^0]+[2^（-2）]*[(2又1/4)^-1/2]-[(0.01)^0.5]](/uploads/image/z/971743-31-3.jpg?t=%E8%AE%A1%E7%AE%97%5B%EF%BC%882%E5%8F%883%2F5%EF%BC%89%5E0%5D%2B%5B2%5E%EF%BC%88-2%EF%BC%89%5D%2A%5B%282%E5%8F%881%2F4%29%5E-1%2F2%5D-%5B%280.01%29%5E0.5%5D)
计算[（2又3/5）^0]+[2^（-2）]*[(2又1/4)^-1/2]-[(0.01)^0.5]
计算[（2又3/5）^0]+[2^（-2）]*[(2又1/4)^-1/2]-[(0.01)^0.5]
计算[（2又3/5）^0]+[2^（-2）]*[(2又1/4)^-1/2]-[(0.01)^0.5]
[（2又3/5）^0]+[2^（-2）]*[(2又1/4)^-1/2]-[(0.01)^0.5]
=1+1/4*(9/4)^-1/2-0.1
=1+1/4*2/3-0.1
=1+1/6-0.1
=11/15
=1+0.25*(2/3)+0.1=1.1+1/6=19/15 ∵一元二次方程有两个不同的实数根∴（2a-8) 2;-4*1*（a 2;+1)>b^2-4ac>0 4(a-4)^