... ∫+−=10 2 65xxdxJ; )2) (3( ) 32 ( )( 23 )2) (3( 1−−+−+=−+−=−−xxBAxBAxBxAxx(0 ,25 )−==⇒−=+=+⇒111 32 0BABABA(0 ,25 )∫ ∫−−=−−−=−−−=⇒1010101010|| 2 3 |ln| |2| ln| |3| ln 23 xxxxxdxxdxJ(0 ,25 )J = 2ln2 – ln3 (0 ,25 )Bài 3: Giải phương trình 2& apos ;2 121 2 131 01 32 iZZ ... x2sin 2 1(0 ,25 ) 2 0 2 0|2cos 2 12sin|2sin. 2 12 ππxxdxxxI =−+=(0,5)1)11( 2 1−=−−=⇒ I(0 ,25 )b) ∫+−=10 2 65xxdxJ; )2) (3( ) 32 ( )( 23 )2) (3( 1−−+−+=−+−=−−xxBAxBAxBxAxx(0 ,25 )−==⇒−=+=+⇒111 32 0BABABA(0 ,25 )∫ ... =+−−−−+−=−−=+=014 )31 (3 )21 (2) 1( 31 21 1ttttztytx(0 ,25 ))4,1,0(1 Ht ⇒−=⇒(0 ,25 )* A’ đối xứng A qua (P) 'HAAH =⇔ hay AHA 2A' =⇒ A’ (-1 ,3, 7) (0 ,25 )c) 1414|14 32 1 |),(=+−+=PAd(0 ,25 )Rc...