DE THI THU DH (MAU) 1 2

... ) u n (1; 1; 1) a b c 0 (1) α⊂ α ⇒ ⊥ = − ⇔ − + =r r Ta coù: 2 2 2 2 2 2a 2b 2c 2 cos(u; u ) 2 a b c . 1 2 2∆+ += =+ + + +r r 2 2 2 2 2 2 2 2 2 2(a 2b 2c) 9(a b c ) 2( a 2a 2c 2c) 9[a ... )40;0ππ=⇒=⇒∈xkxKL:0 ,25 0 .25 CâuIII (1. 0điểm)( ) 1 1 2 26 60 0 1 13 2 2 2 0 0 2 x 0 t 0t.2tdt t dtÑaët t x t x 2tdt dx Vaäy I 2 x 1 t 1 t 1 t 1 dut 0 u 0 2 1 3Ñaët u t du 3t dt I 2 dut 1 u 1 u 1 3 1 uu ... +( ) 2 1 1 2 ' 2x x k = =( ) ( ) 2 1 2 1 2 2y y k x x k k = = Do ú : Theo gi thit BC= 2 23 34 4 2 2 4 4 8 0 1k k k k k + = + = =Vy :y=x +2 0 .25 0 .25 0 .25 Cõu II (2. 0im) 1. ĐK:...

... 2 2 1x ux u x    0 ,25 2 2 1 1 1 1 2 2ux dx duu u        Đổi cận 1x   thì 2 1u   ; 1x  thì 2 1. u   0 ,25 2 1 2 1 2 1 2 2 2 1 2 1 2 1 1 1 1 1 1 2 1 ...    0 ,25    2 2sin 2 3sin 2 1 3cos2 2sin 2 1 0x x x x        2sin 2 1 sin 2 3cos2 1 0x x x     0 ,25   12 1 5sin 2 2 12 , , , . 1 sin 2 3 2 47. 12 x kxx lk ... 1 2 1 2 1 2 (1 )dudu duuIu u u u               0 ,25 3 (1, 0 điểm) = 2 1 2 1 2 2 1 2 1 1 1 1 1 1 2 1 2 1 duduu u u u         =1...