计算[(2又3⼀5)^0]+[2^(-2)]*[(2又1⼀4)^-1⼀2]-[(0.01)^0.5]

2024-12-17 17:18:20
推荐回答(2个)
回答1:

[(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

回答2:

=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)^