... 0 ,25 0 ,25 0, 0,5 0,5 0 ,25 0 ,25 0,5 ( a2 + b2 – – 4b) + [a( b – 2) – a( b + 2) + 4a] ⇔ i = a2 + b2 – 4b – = Ta lại có:= Vậy môđun z – 2i Câu 7b Câu 8b z − 2i a= + b (b4b 2) i4 = a= + 2= 2 b2+ ... −coscos−+= 1cos2 cos x ) π 2 x.( (2 + dx x2 t= ∫ cos11=x2.3((1⇒dt cosA'x ) = = = = Vậy I = 1 + 22 2 2 ( ln ln 2 ) − ln − ln 121 ∫1 3.(1t+ t Theo góc ∠A' AH 60 a = 2t 2 ln2 t ) dt 2t − + ... C2 ) nên ta có: Ta có: ; ⇒II2II= I=II1 (21 − aR2a 1− 116)5 = − = R ; 2a − = I 1 −− I− a ; +2 R R 5) ( 2 I =( + R ) R I I − I I = 2 ( + a )I 2+ ⇔a −1 62 − a + ( 2a − 5) = 5a − 60a + 26 0a = 25 a=...