... 2 (1, 0) + 12 ,(),()M Ndd∈ nên ta giả sử 11 1 22 2 1 2 12 12 (;;2), (1 2; ;1 ) ( 2 1; ;2 1) MtttN tt t NM t t tttt−− + ⇒ = + + − − − JJJJG . + MN song song mp(P) nên: nN 1 2 12 12 . 0 1. ( 2 1) 1. ( ... nN 1 2 12 12 . 0 1. ( 2 1) 1. ( ) 1( 2 1) 0 P M t t t t t t= ⇔++−−+−−= JJG JJJJG ) 21 1 11 (1; 2;31ttNMt tt⇔=−⇒ =−+ − JJJJG . + Ta có: 1 22 2 2 11 1 11 1 0 2( 1) (2)( 31) 2740 4 7 t MN t t t t t t = ⎡ ... có: 222 11 111 11 () 1( )1( )1( )3 3 ab bc ca a b c b c a c a b c b c abc abc 1+ + ++≤++== ++ ++ ++ , . Dấu “=” xảy ẩ khi và chỉ khi 1, 3 1, ( , , 0).abc ab bc ca a b c a b c= ++=⇒=== > 1, 0 Câu...