... mmaDkaDkxx 5,722 1 1 21 ==↔=λλ5,7 1 5,2) .1, 0( 1 5,2. 1 2 1 1=+=↔λλkk=+=↔=+=↔3 )1, 0(35,75,2) .1, 0(5,2. 12 11 12 11 λλλλkkkkSuy ra 30 1, 0)(33 )1, 03(3 )1, 03( 21 212 1 211 12 1 2kkkkkkkkkkkkk=−↔=−↔=+↔=+Do ... JUechAUeAchhALhAl 19 199834 1 1 1 1 10 .5 21, 505 ,1. 10.6 ,1 10.276 10 .3 .10 .625,6−−−−=−=−=↔+=λλ và JUechAUeAchhCuhCU 19 1998342222 10 .638,686,0 .10 .6 ,1 10.248 10 .3 .10 .625,6−−−−=−=−=↔+=λλVậy ta có VeAchUUeAchALhAl558 ,1 10.6 ,1 10.5 21, 5 10 .248 10 .3 .10 .625,6 19 19 983422=−=−=↔+=−−−−λλCâu ... D. 10 sàGia:Khi u 1 =1, 2mV thì i 1 =1, 8mA ta có: 202 1 2 1 2 1 2 1 2 1 LILicu =+Khi u2=-0,9mV thì i2=2,4mA ta có: 2022222 1 2 1 2 1 LILicu =+=>Ta có 222 1 2 1 222 1 22222 1 22222 1 2 1 )()()(2 1 2 1 2 1 2 1 uuiiLCiiLuucLicuLicu−−=↔−=−↔+=+FC5232323236 10 .2 )10 .9,0( )10 .2 ,1( ] )10 .8 ,1( )10 .4,2[ (10 .5−−−−−−=−−=↔Chu...