二次根式中考题都在这里了

A.1和2 B.2和3 C.3和4 D.4和5 7 16.已知:a、b为两个连续的整数,且a<15

D

6.下列式子运算正确的是( ) A.3?2?1 B.8?42 C. 13?3 D.12?3?12?3?1 B 7.下列计算正确的是( ) A.3+3=6 B.3-3=0 C.3·3=9 D.(-3)2=-3 8.计算:2?631 ?1=_______. 9.计算:12?3?______. 10.已知20n是整数,则满足条件的最小正整数n为( ) A.2 B.3 C.4 D.5 33 D 11计算:18?8?___________. 12.化简:18?8=________. 13.计算:(π?1)?12??3 解:(π?1)?12??3?1?23?3?1?3. 14.计算:(π?2009)0?12?|3?2| 解:原式?1?23?2?3?(1?2)?(23?3)?3?3 15.在算式()□()的□中填上运算符号,使结果最大,这个运算002 2 D 符号是( ) A.加号 B.减号 C.乘号 D.除号 ??1312?2?48?23. 16.计算:????3??解:原式??63? ??28142?3?23?3?43??23?. 333?D 17.若x?m?n,y?m?n,则xy的值是( ) A.2m B.2n C.m?n D.m?n 10

18. 计算147?75+27之值为何? A.53 B.33 C.311 D. 911 A 954319. 17.计算之值为何? ??12612A.B 33333 B. C. D. 12634 1120.计算:8??2?______. 32答案:2?13 3B 21. 计算1925?4之值为何? 16365577(A) 2(B) 3(C) 4 (D) 5121212 12 提示:带分数化为假分数0122.计算:(?3)?27?1?2?. 3?2解:(?3)?27?1?2??10 1=1?33?2?1?3?2=?23. 3?2 ?1?23.计算:12????2sin60? ?2?解:原式=23?2?2?3 =23?2?3=3?2 2?1?1?24.计算:|?2|????sin45°?(2009)0 ?2?解:原式? 2?2?2?1?1 2 第八类:先化简,再求值 主要结合分式的化简,涉及较少的二次根式的综合运算, 11

是中考的热点。 2011,则m5?2m4?2011m3的值是______. 2012?120112012?1提示:m?==2012+1, 2012?12012?1则m5?2m4?2011m3=m3(m2?2m?2011)?m3[(m?1)2?2012)]?0. 1. 若m?2.先化简,再求值:0 x?4x?4??x2?2x?,其中x=2. x?22 (x?2)21112解:当x=2时,原式?. ????x?2x(x?2)x223.先化简,再求值: x?x2?1?3x,其中x?2?2 ????x?x?1x?1??3x?x?1?x?x?1??x2?1解法一:原式=????x ????????x?1x?1x?1x?1??2x2?4xx2?13x2?3x?x2?xx2?1 = = ???x?1??x?1?x?x?1??x?1?x=2x?x?2??x?1??x?1??=2?x?2? ?x?1??x?1?x 当x?2?2时,原式=22?2?2=22 ??3xx2?1xx2?1???解法二:原式= x?1xx?1x =3x?x?1??x?1?x?x?1??x?1???? x?1xx?1x =3?x?1???x?1?=3x?3?x?1=2x?4 当x?2?2时,原式=(22?2)?4=22 5x?3)?,其中x=2?3. x?22x?4 4.先化简,再求值:(x?2?x2?452(x?2)?)?解: 原式=( x?2x?2x?3 12

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