... = uuuur uuuur 2 2 2 1 < /b> a 3a a BM.CB 0 4 4 2 ⇒ BM ⊥ B 1 < /b> C Ta có 1 < /b> B. B (0,0,a)= uuuuur ⇒ = = uuuur uuuur uuuur uuuur uuuur 1 < /b> 1 1 < /b> 1 [BM .B C].BB a 30 d(BM ,B C) 10< /b> [BM .B C] x C B ----------@--------- ... (1 < /b> – tgx) (1 < /b> + sin2x) = 1 < /b> + tgx (1)< /b> Đặt: t = tgx 2 t1 t2 x2sin + =⇒ . Pt (1)< /b> thành ( ) 2 2t 1 < /b> t 1 < /b> 1 t 1 < /b> t − + = + ÷ + ( ) ( ) 2 2 1 < /b> t t 1 < /b> (t 1)< /b> (1 < /b> t )⇔ − + = + + ( ) ( ) 2 t 1 < /b> 0 hay 1 < /b> ... ( ) 2 4 41 < /b> 1t22t332t 21 < /b> = ++ −+−−+ 6 12< /b> t 6 6 12< /b> t 6 hay 6 12< /b> t 6 t 1hay t 0⇔ − + = ⇔ − + = − + = − ⇔ = = . t = 1 < /b> ⇒ t' = 1 < /b> ⇒ M 1 < /b> (3, 0, 2) N 1 < /b> ( 1,< /b> –4, 0) . t = 0 ⇒ t' = 0 ⇒ M 2 (1,< /b> 3, 0)...