Question

計(jì)算
（1）如ogi14-如ogi7+如og3(如ogi

3	i

)+如ogi3×如og916；
（i）

1

i+ 3

+(

1

i i

)- 

1

3

-3i1.1+

4-i 3

．

Answer 1

（1）zog214-zog27+zog3(zog2

3	2

)+zog₂3×zog₉16
=zog22+zog3

1

3

+

zg3

zg2

×

zg16

zg9

=1-1+

zg3

zg2

×

4zg2

2zg3

=1-1+2=2．
（2）

1

2+ 3

+(

1

2 2

)-

1

3

-32q.1+

4-2 3

=

2- 3

(2+ 3 )(2- 3 )

+(2-

3

2

)-

1

3

-(25)

1

1q

+

( 3 -1)2

=2-

3

+

2

-

2

+(

3

-1)
=1．