Remove members in Address list in exchange 2010

how to remove members in certain address list in exchange 2010.i have created many equipments mailboxesnow they are under All rooms and at the same time they are under laptop address list(created).how can i remove them from All Rooms address list to keep in it only rooms. laptops address list All Rooms address list
alaayehyaAsked:
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Hendrik WieseInformation Security ManagerCommented:
I have not tested the script but try running the following in EMS.

Set-AddressList -Identity "All Rooms" -RecipientFilter {(Alias -ne $null -and (RecipientDisplayType -eq 'ConferenceRoomMailbox' -or RecipientDisplayType -eq
'SyncedConferenceRoomMailbox'))}

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Hendrik WieseInformation Security ManagerCommented:
After you ran the script you can follow my article to manually update your GAL: Manually Update Global Address List (GAL) on Exchange 2010

Please mark the article as helpful should it be part of the solution?

Cheers!!!
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alaayehyaAuthor Commented:
this script will clear laptops and remove them from ALL ROOM Address list?
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Making Bulk Changes to Active Directory

Watch this video to see how easy it is to make mass changes to Active Directory from an external text file without using complicated scripts.

Hendrik WieseInformation Security ManagerCommented:
It will change the query for the All Rooms that is loaded to not include equipment mailboxes.
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alaayehyaAuthor Commented:
now i am doing this script but nothing is being changed it seems it is not executed should I put any more parameters?
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Hendrik WieseInformation Security ManagerCommented:
Am currently on mobile, Will revise as soon as I get behind a laptop.
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e_aravindCommented:
When you Open-up the "Exchange Management Console" > Organization Configuration > Mailbox > Address Lists

Choose the "All Rooms" > Next

The "Recipients filter" should have the below value:
((Alias -ne $null) -and (((RecipientDisplayType -eq 'ConferenceRoomMailbox') -or (RecipientDisplayType -eq 'SyncedConferenceRoomMailbox'))))

you can use the Preview to see the result which would be listed for the "All Rooms"
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alaayehyaAuthor Commented:
partially solved
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