1、2 本讲聚焦 Highlights 1.2.3.4.5.考点评析 Analysis 知识精讲 Concentration 1Conception 1.13,1.6,+,+2+131.6,2,2.3.110 3.052+124()A1 B2 C3 D4 3 Exercise(1)3122.40 23()A1 B2 C3 D4(2)()A5B“”CDa2(1)“”()AB100 100 C2 2 D(2)701+701256()A256 B256 C957D445 Exercise(1)30+30 10 _(2)5050+100()A50 B50 C100 D100(3)()A3 3 B3 C3
2、CCD65:6060:653(1)4.003.850.154.22_.4(2)“250.25”()A25.28 B25.18 C24.69 D24.25 Exercise(1)4519099 8:1519:451+7:30()A2+B2C1.50D7.30(2)(203)C()A17 C 20 C B17 C 20 C C17 C 23 C D17 C 24 C 2Conception 1.Oorigin unit length positive direction number axis 2.3.4.aa_a_4()A.0 原点 单位长度 5 B.C.D.Exercise()A BCD5AB
3、C D 3,ABCDC4 ABCDAB C Exercise(1)A()A1.5 B1.5 C2.6 D2.6 0 1 2 3 A 01 01 0 01 2 6(2)pq pq；p0；pq；qq.6352AB.(1)A3A7BB,AB.(2)A3A75BB,AB.Exercise 5()A13B13C14D14q 0 p 7 1.，opposite number .2.3.a=_ a4.0=7122+_ _ 05+_;_2.374_;()ab_(+)a b _Exercise 5a=a=_()a=_()a+8a=1()a=_ 0a a _ a=a a=_ a=8.3,a=_abab+_8()A
4、“+”“”3Conception 8 BCD“”Exercise()A0BCD9(1)4 m1m=_(2)0mn+=0np+=0mq=0mnpq()ApqBmpCmnDmpExercise m n2223mnmn+=_.10.(1)(2)+;(2)(2015);(3)(18)+;(4)2 ()3+.Exercise(1)1(2)3;(2)(5)+;(3)(0.25);9(4)1()2+;(5)(1)+;(6)()a.4Conception 1.2.a 3.0(=0)(0)(0)4.a0a 0a=0a|_a=|_a=|_a=11(1)122；0.8；0；1(3)9.(2)9_,4Exercise
5、(1)2_2_(6)_|8|_|4|_ 0 0 0 10(2)|3x=x=_|3|0 x=x=_.12(1)()A|a BC|=|ababD(2)()A|4|4B|4|4C|4|4 4 DExercise 1()|b|;|p|=qpq;|p|+p=0pA1B2C3D413(1)m|mm=m_.(2)|mn=mn_.(3)|2|2aa=aAa0 Ba0 Ca 0 Da0 Exercise(1)|4a=a=_|2|a=a=_|3|2a=a=_.11(2)|2|2aa=()A2a=B2a C2aD0a=14|x_.Exercise|4|1x=_1.x_.15Aa,Bb2|5|(1)0ab+=,AB|ABab=AB|AB.Exerciseab|3|2|0ab+=,a+b.51.2.16(1)1 _|0|(2)1 _|3|(3)1()4 _ 1|2(4)|1|_|0.2|.Exercise(1)5|3|6_ 7|3|8(2)|3.2|_(3.2)(3)|0.0001|_ 2000(4)|1.38|_|1.3846|(5)|_|3.14|.
