Question

求值：①2log₃9+log₂

1

4

-0.7⁰-2^-1+25

1

2

=______；
②sin²120°+cos180°+tan45°-cos²-（-330°）+sin（-210°）=______．

Answer 1

①2log₃9+log₂

1

4

-0.7⁰-2^-1+25

1

2

=4log33-2log22-1-

1

2

+5=4-2-1-

1

2

+5
=5

1

2

②sin²120°+cos180°+tan45°-cos²-（-330°）+sin（-210°）
=（

3

2

）²-1+1-cos²30°-sin210°
=

3

4

-（

3

2

）²+sin30°=sin30°
=

1

2