Question

高一数学计算题要过程

Answer 1

6、
（1）原式=1.1^0+4^(3/3)-1/0.25+lg25+lg2^2=1+4-4+lg(25*4)=1+2=3
（2）原式=sin²(-1)+cos²1-sin(5π/6-2π)cos(5π-π/3)=(-sin1)²+cos²1-sin(5π/6)*(-cosπ/3)
=sin²1+cos²1-(1/2)*(-1/2)=1+1/4=5/4

Answer 2

Answer 3

1. 原式=1.1^0+4-4+lg25+lg4

         =1+lg25*4

         =1+2

         =3

2. 原式 =sin^2(-360-60)+(根号3/2)^2-sin(-360+150)cos(720+120)

          =(sin60)^2+3/4-sin150cos120

         =3/4+3/4-1/2*(-1/2)

         =7/4

Answer 4

1+4-4+2=3
(-二分之根号三)^2+（二分之根号三)^2-0.5*（-0.5）=41/36

Answer 5

(1)原式=1.1的0次方+4-0.25+lg25+lg2的2次方=1+4-0.25+lg25+lg4=1+4-0.25+lg100=1+4-0.25+2=6.75