Description
π 6) ! -1
β 2 + = sin ( )
β =
+ 2 =
(1) (5)
(1) = = 1, (5) = β1, = 0, (1) = 1
π 7) 1 5 – ./ ,! ( ) ! * + ( β -2
1
= + β + exp (β )
(3, )
(1, ) = sin(2″ ) = 0 (, 0) = 1


π 6) ! -1
β 2 + = sin ( )
β =
+ 2 =
(1) (5)
(1) = = 1, (5) = β1, = 0, (1) = 1
π 7) 1 5 – ./ ,! ( ) ! * + ( β -2
1
= + β + exp (β )
(3, )
(1, ) = sin(2″ ) = 0 (, 0) = 1
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