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2021年福建专用高考数学总复习第七章不等式推理与证明课时规范练35直接证明与间接证明理新人教A版.pdf

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《2021年福建专用高考数学总复习第七章不等式推理与证明课时规范练35直接证明与间接证明理新人教A版.pdf》由会员分享，可在线阅读，更多相关《2021年福建专用高考数学总复习第七章不等式推理与证明课时规范练35直接证明与间接证明理新人教A版.pdf（5页珍藏版）》请在淘文阁 - 分享文档赚钱的网站上搜索。

1、1/5 课时规范练 35 直接证明与间接证明一、基础巩固组1.要证:a2+b2-1-a2b20,只需证明()A.2ab-1-a2b20B.a2+b2-1-0C.-1-a2b20D.(a2-1)(b2-1)02.用反证法证明结论“三角形内角至少有一个不大于60”,应假设()A.三个内角至多有一个大于60B.三个内角都不大于 60C.三个内角都大于60D.三个内角至多有两个大于603.(2017 河南郑州模拟)设x0,P=2x+2-x,Q=(sin x+cos x)2,则()A.PQB.Pb0,m=,n=,则m,n的大小关系是.6.设a,b,c均为正数,且a+b+c=1,求证:ab+bc+ac.|

2、精.|品.|可.|编.|辑.|学.|习.|资.|料.*|*|*|*|欢.|迎.|下.|载.第 1 页，共 5 页2/5 7.(2017 河北唐山模拟)已知a0,1,求证:.二、综合提升组8.设f(x)是定义在R 上的奇函数,且当x0 时,f(x)单调递减,若x1+x20,则f(x1)+f(x2)的值()A.恒为负值B.恒等于零C.恒为正值D.无法确定正负9.如果A1B1C1的三个内角的余弦值分别等于A2B2C2的三个内角的正弦值,则()A.A1B1C1和A2B2C2都是锐角三角形B.A1B1C1和A2B2C2都是钝角三角形C.A1B1C1是钝角三角形,A2B2C2是锐角三角形D.A1B1C1是

3、锐角三角形,A2B2C2是钝角三角形?导学号 21500552?10.已知a,b是不相等的正数,x=,y=,则x,y的大小关系是.11.已知函数f(x)=ln(1+x),g(x)=a+bx-x2+x3,函数y=f(x)与函数y=g(x)的图象在交点(0,0)处有公共切线.(1)求a,b的值;(2)证明f(x)g(x).|精.|品.|可.|编.|辑.|学.|习.|资.|料.*|*|*|*|欢.|迎.|下.|载.第 2 页，共 5 页文档编码:CV2L6B4B7Q8 HY7H8O7H2S6 ZU3L1S7E7E6文档编码:CV2L6B4B7Q8 HY7H8O7H2S6 ZU3L1S7E7E6文档编

4、码:CV2L6B4B7Q8 HY7H8O7H2S6 ZU3L1S7E7E6文档编码:CV2L6B4B7Q8 HY7H8O7H2S6 ZU3L1S7E7E6文档编码:CV2L6B4B7Q8 HY7H8O7H2S6 ZU3L1S7E7E6文档编码:CV2L6B4B7Q8 HY7H8O7H2S6 ZU3L1S7E7E6文档编码:CV2L6B4B7Q8 HY7H8O7H2S6 ZU3L1S7E7E6文档编码:CV2L6B4B7Q8 HY7H8O7H2S6 ZU3L1S7E7E6文档编码:CV2L6B4B7Q8 HY7H8O7H2S6 ZU3L1S7E7E6文档编码:CV2L6B4B7Q8 HY7H8O

5、7H2S6 ZU3L1S7E7E6文档编码:CV2L6B4B7Q8 HY7H8O7H2S6 ZU3L1S7E7E6文档编码:CV2L6B4B7Q8 HY7H8O7H2S6 ZU3L1S7E7E6文档编码:CV2L6B4B7Q8 HY7H8O7H2S6 ZU3L1S7E7E6文档编码:CV2L6B4B7Q8 HY7H8O7H2S6 ZU3L1S7E7E6文档编码:CV2L6B4B7Q8 HY7H8O7H2S6 ZU3L1S7E7E6文档编码:CV2L6B4B7Q8 HY7H8O7H2S6 ZU3L1S7E7E6文档编码:CV2L6B4B7Q8 HY7H8O7H2S6 ZU3L1S7E7E6文档编

6、码:CV2L6B4B7Q8 HY7H8O7H2S6 ZU3L1S7E7E6文档编码:CV2L6B4B7Q8 HY7H8O7H2S6 ZU3L1S7E7E6文档编码:CV2L6B4B7Q8 HY7H8O7H2S6 ZU3L1S7E7E6文档编码:CV2L6B4B7Q8 HY7H8O7H2S6 ZU3L1S7E7E6文档编码:CV2L6B4B7Q8 HY7H8O7H2S6 ZU3L1S7E7E6文档编码:CV2L6B4B7Q8 HY7H8O7H2S6 ZU3L1S7E7E6文档编码:CV2L6B4B7Q8 HY7H8O7H2S6 ZU3L1S7E7E6文档编码:CV2L6B4B7Q8 HY7H8O

7、7H2S6 ZU3L1S7E7E6文档编码:CV2L6B4B7Q8 HY7H8O7H2S6 ZU3L1S7E7E6文档编码:CV2L6B4B7Q8 HY7H8O7H2S6 ZU3L1S7E7E6文档编码:CV2L6B4B7Q8 HY7H8O7H2S6 ZU3L1S7E7E6文档编码:CV2L6B4B7Q8 HY7H8O7H2S6 ZU3L1S7E7E6文档编码:CV2L6B4B7Q8 HY7H8O7H2S6 ZU3L1S7E7E6文档编码:CV2L6B4B7Q8 HY7H8O7H2S6 ZU3L1S7E7E6文档编码:CV2L6B4B7Q8 HY7H8O7H2S6 ZU3L1S7E7E6文档编

8、码:CV2L6B4B7Q8 HY7H8O7H2S6 ZU3L1S7E7E6文档编码:CV2L6B4B7Q8 HY7H8O7H2S6 ZU3L1S7E7E6文档编码:CV2L6B4B7Q8 HY7H8O7H2S6 ZU3L1S7E7E6文档编码:CV2L6B4B7Q8 HY7H8O7H2S6 ZU3L1S7E7E6文档编码:CV2L6B4B7Q8 HY7H8O7H2S6 ZU3L1S7E7E6文档编码:CV2L6B4B7Q8 HY7H8O7H2S6 ZU3L1S7E7E6文档编码:CV2L6B4B7Q8 HY7H8O7H2S6 ZU3L1S7E7E6文档编码:CV2L6B4B7Q8 HY7H8O

9、7H2S6 ZU3L1S7E7E6文档编码:CV2L6B4B7Q8 HY7H8O7H2S6 ZU3L1S7E7E6文档编码:CV2L6B4B7Q8 HY7H8O7H2S6 ZU3L1S7E7E6文档编码:CV2L6B4B7Q8 HY7H8O7H2S6 ZU3L1S7E7E6文档编码:CV2L6B4B7Q8 HY7H8O7H2S6 ZU3L1S7E7E6文档编码:CV2L6B4B7Q8 HY7H8O7H2S6 ZU3L1S7E7E6文档编码:CV2L6B4B7Q8 HY7H8O7H2S6 ZU3L1S7E7E6文档编码:CV2L6B4B7Q8 HY7H8O7H2S6 ZU3L1S7E7E6文档编

10、码:CV2L6B4B7Q8 HY7H8O7H2S6 ZU3L1S7E7E63/5 三、创新应用组12.(2017 贵州安顺调研)已知函数f(x)=3x-2x,求证:对于任意的x1,x2R,均有f.13.在等差数列 an 中,a1=3,其前n项和为Sn,等比数列 bn的各项均为正数,b1=1,公比为q(q1),且b2+S2=12,q=.(1)求an与bn;(2)证明:+.?导学号 21500553?课时规范练35直接证明与间接证明1.D在各选项中,只有(a2-1)(b2-1)0?a2+b2-1-a2b20,故选 D.2.C“三角形内角至少有一个不大于60”即“三个内角至少有一个小于等于60”,其

11、否定为“三角形内角都大于60”.故选 C.3.A因为 2x+2-x2=2(当且仅当x=0 时等号成立),而x0,所以P2;又(sin x+cos x)2=1+sin 2x,而 sin 2x1,所以Q2.于是PQ.故选 A.4.Da0,b0,c0,6,当且仅当a=b=c=1 时,等号成立,故三者不能都小于2,即至少有一个不小于2.5.mn 方法一:(取特殊值法)取a=2,b=1,得mn.方法二:(分析法)a0,显然成立.6.证明由a2+b22ab,b2+c22bc,c2+a22ac得a2+b2+c2ab+bc+ca.由题设得(a+b+c)2=1,即a2+b2+c2+2ab+2bc+2ca=1.|

12、精.|品.|可.|编.|辑.|学.|习.|资.|料.*|*|*|*|欢.|迎.|下.|载.第 3 页，共 5 页文档编码:CV2L6B4B7Q8 HY7H8O7H2S6 ZU3L1S7E7E6文档编码:CV2L6B4B7Q8 HY7H8O7H2S6 ZU3L1S7E7E6文档编码:CV2L6B4B7Q8 HY7H8O7H2S6 ZU3L1S7E7E6文档编码:CV2L6B4B7Q8 HY7H8O7H2S6 ZU3L1S7E7E6文档编码:CV2L6B4B7Q8 HY7H8O7H2S6 ZU3L1S7E7E6文档编码:CV2L6B4B7Q8 HY7H8O7H2S6 ZU3L1S7E7E6文档编码

13、:CV2L6B4B7Q8 HY7H8O7H2S6 ZU3L1S7E7E6文档编码:CV2L6B4B7Q8 HY7H8O7H2S6 ZU3L1S7E7E6文档编码:CV2L6B4B7Q8 HY7H8O7H2S6 ZU3L1S7E7E6文档编码:CV2L6B4B7Q8 HY7H8O7H2S6 ZU3L1S7E7E6文档编码:CV2L6B4B7Q8 HY7H8O7H2S6 ZU3L1S7E7E6文档编码:CV2L6B4B7Q8 HY7H8O7H2S6 ZU3L1S7E7E6文档编码:CV2L6B4B7Q8 HY7H8O7H2S6 ZU3L1S7E7E6文档编码:CV2L6B4B7Q8 HY7H8O7

14、H2S6 ZU3L1S7E7E6文档编码:CV2L6B4B7Q8 HY7H8O7H2S6 ZU3L1S7E7E6文档编码:CV2L6B4B7Q8 HY7H8O7H2S6 ZU3L1S7E7E6文档编码:CV2L6B4B7Q8 HY7H8O7H2S6 ZU3L1S7E7E6文档编码:CV2L6B4B7Q8 HY7H8O7H2S6 ZU3L1S7E7E6文档编码:CV2L6B4B7Q8 HY7H8O7H2S6 ZU3L1S7E7E6文档编码:CV2L6B4B7Q8 HY7H8O7H2S6 ZU3L1S7E7E6文档编码:CV2L6B4B7Q8 HY7H8O7H2S6 ZU3L1S7E7E6文档编码

15、:CV2L6B4B7Q8 HY7H8O7H2S6 ZU3L1S7E7E6文档编码:CV2L6B4B7Q8 HY7H8O7H2S6 ZU3L1S7E7E6文档编码:CV2L6B4B7Q8 HY7H8O7H2S6 ZU3L1S7E7E6文档编码:CV2L6B4B7Q8 HY7H8O7H2S6 ZU3L1S7E7E6文档编码:CV2L6B4B7Q8 HY7H8O7H2S6 ZU3L1S7E7E6文档编码:CV2L6B4B7Q8 HY7H8O7H2S6 ZU3L1S7E7E6文档编码:CV2L6B4B7Q8 HY7H8O7H2S6 ZU3L1S7E7E6文档编码:CV2L6B4B7Q8 HY7H8O7

16、H2S6 ZU3L1S7E7E6文档编码:CV2L6B4B7Q8 HY7H8O7H2S6 ZU3L1S7E7E6文档编码:CV2L6B4B7Q8 HY7H8O7H2S6 ZU3L1S7E7E6文档编码:CV2L6B4B7Q8 HY7H8O7H2S6 ZU3L1S7E7E6文档编码:CV2L6B4B7Q8 HY7H8O7H2S6 ZU3L1S7E7E6文档编码:CV2L6B4B7Q8 HY7H8O7H2S6 ZU3L1S7E7E6文档编码:CV2L6B4B7Q8 HY7H8O7H2S6 ZU3L1S7E7E6文档编码:CV2L6B4B7Q8 HY7H8O7H2S6 ZU3L1S7E7E6文档编码

17、:CV2L6B4B7Q8 HY7H8O7H2S6 ZU3L1S7E7E6文档编码:CV2L6B4B7Q8 HY7H8O7H2S6 ZU3L1S7E7E6文档编码:CV2L6B4B7Q8 HY7H8O7H2S6 ZU3L1S7E7E6文档编码:CV2L6B4B7Q8 HY7H8O7H2S6 ZU3L1S7E7E6文档编码:CV2L6B4B7Q8 HY7H8O7H2S6 ZU3L1S7E7E6文档编码:CV2L6B4B7Q8 HY7H8O7H2S6 ZU3L1S7E7E6文档编码:CV2L6B4B7Q8 HY7H8O7H2S6 ZU3L1S7E7E6文档编码:CV2L6B4B7Q8 HY7H8O7

18、H2S6 ZU3L1S7E7E6文档编码:CV2L6B4B7Q8 HY7H8O7H2S6 ZU3L1S7E7E6文档编码:CV2L6B4B7Q8 HY7H8O7H2S6 ZU3L1S7E7E6文档编码:CV2L6B4B7Q8 HY7H8O7H2S6 ZU3L1S7E7E6文档编码:CV2L6B4B7Q8 HY7H8O7H2S6 ZU3L1S7E7E64/5 所以 3(ab+bc+ca)1,即ab+bc+ca7.证明由已知1 及a0 可知 0b1,只需证 1+a-b-ab1,只需证a-b-ab 0,即1,即1,这是已知条件,所以原不等式得证.8.A由f(x)是定义在R上的奇函数,且当x0 时,f

19、(x)单调递减,可知f(x)是 R上的单调递减函数.由x1+x20,可知x1-x2,即f(x1)f(-x2)=-f(x2),则f(x1)+f(x2)0,故选 A.9.D由条件知,A1B1C1的三个内角的余弦值均大于 0,则A1B1C1是锐角三角形,且A2B2C2不可能是直角三角形.假设A2B2C2是锐角三角形.由得则A2+B2+C2=,这与三角形内角和为180相矛盾.因此假设不成立,故A2B2C2是钝角三角形.10.x22(a+b)a+b+2a+b,即x-1).h(x)=-x2+x-1=,h(x)在(-1,0)内为增函数,在(0,+)内为减函数.h(x)max=h(0)=0,即h(x)h(0

20、)=0,即f(x)g(x).|精.|品.|可.|编.|辑.|学.|习.|资.|料.*|*|*|*|欢.|迎.|下.|载.第 4 页，共 5 页文档编码:CV2L6B4B7Q8 HY7H8O7H2S6 ZU3L1S7E7E6文档编码:CV2L6B4B7Q8 HY7H8O7H2S6 ZU3L1S7E7E6文档编码:CV2L6B4B7Q8 HY7H8O7H2S6 ZU3L1S7E7E6文档编码:CV2L6B4B7Q8 HY7H8O7H2S6 ZU3L1S7E7E6文档编码:CV2L6B4B7Q8 HY7H8O7H2S6 ZU3L1S7E7E6文档编码:CV2L6B4B7Q8 HY7H8O7H2S6

21、ZU3L1S7E7E6文档编码:CV2L6B4B7Q8 HY7H8O7H2S6 ZU3L1S7E7E6文档编码:CV2L6B4B7Q8 HY7H8O7H2S6 ZU3L1S7E7E6文档编码:CV2L6B4B7Q8 HY7H8O7H2S6 ZU3L1S7E7E6文档编码:CV2L6B4B7Q8 HY7H8O7H2S6 ZU3L1S7E7E6文档编码:CV2L6B4B7Q8 HY7H8O7H2S6 ZU3L1S7E7E6文档编码:CV2L6B4B7Q8 HY7H8O7H2S6 ZU3L1S7E7E6文档编码:CV2L6B4B7Q8 HY7H8O7H2S6 ZU3L1S7E7E6文档编码:CV2L

22、6B4B7Q8 HY7H8O7H2S6 ZU3L1S7E7E6文档编码:CV2L6B4B7Q8 HY7H8O7H2S6 ZU3L1S7E7E6文档编码:CV2L6B4B7Q8 HY7H8O7H2S6 ZU3L1S7E7E6文档编码:CV2L6B4B7Q8 HY7H8O7H2S6 ZU3L1S7E7E6文档编码:CV2L6B4B7Q8 HY7H8O7H2S6 ZU3L1S7E7E6文档编码:CV2L6B4B7Q8 HY7H8O7H2S6 ZU3L1S7E7E6文档编码:CV2L6B4B7Q8 HY7H8O7H2S6 ZU3L1S7E7E6文档编码:CV2L6B4B7Q8 HY7H8O7H2S6

23、ZU3L1S7E7E6文档编码:CV2L6B4B7Q8 HY7H8O7H2S6 ZU3L1S7E7E6文档编码:CV2L6B4B7Q8 HY7H8O7H2S6 ZU3L1S7E7E6文档编码:CV2L6B4B7Q8 HY7H8O7H2S6 ZU3L1S7E7E6文档编码:CV2L6B4B7Q8 HY7H8O7H2S6 ZU3L1S7E7E6文档编码:CV2L6B4B7Q8 HY7H8O7H2S6 ZU3L1S7E7E6文档编码:CV2L6B4B7Q8 HY7H8O7H2S6 ZU3L1S7E7E6文档编码:CV2L6B4B7Q8 HY7H8O7H2S6 ZU3L1S7E7E6文档编码:CV2L

24、6B4B7Q8 HY7H8O7H2S6 ZU3L1S7E7E6文档编码:CV2L6B4B7Q8 HY7H8O7H2S6 ZU3L1S7E7E6文档编码:CV2L6B4B7Q8 HY7H8O7H2S6 ZU3L1S7E7E6文档编码:CV2L6B4B7Q8 HY7H8O7H2S6 ZU3L1S7E7E6文档编码:CV2L6B4B7Q8 HY7H8O7H2S6 ZU3L1S7E7E6文档编码:CV2L6B4B7Q8 HY7H8O7H2S6 ZU3L1S7E7E6文档编码:CV2L6B4B7Q8 HY7H8O7H2S6 ZU3L1S7E7E6文档编码:CV2L6B4B7Q8 HY7H8O7H2S6

25、ZU3L1S7E7E6文档编码:CV2L6B4B7Q8 HY7H8O7H2S6 ZU3L1S7E7E6文档编码:CV2L6B4B7Q8 HY7H8O7H2S6 ZU3L1S7E7E6文档编码:CV2L6B4B7Q8 HY7H8O7H2S6 ZU3L1S7E7E6文档编码:CV2L6B4B7Q8 HY7H8O7H2S6 ZU3L1S7E7E6文档编码:CV2L6B4B7Q8 HY7H8O7H2S6 ZU3L1S7E7E6文档编码:CV2L6B4B7Q8 HY7H8O7H2S6 ZU3L1S7E7E6文档编码:CV2L6B4B7Q8 HY7H8O7H2S6 ZU3L1S7E7E6文档编码:CV2L

26、6B4B7Q8 HY7H8O7H2S6 ZU3L1S7E7E6文档编码:CV2L6B4B7Q8 HY7H8O7H2S6 ZU3L1S7E7E6文档编码:CV2L6B4B7Q8 HY7H8O7H2S6 ZU3L1S7E7E6文档编码:CV2L6B4B7Q8 HY7H8O7H2S6 ZU3L1S7E7E6文档编码:CV2L6B4B7Q8 HY7H8O7H2S6 ZU3L1S7E7E65/5 12.证明要证f,即证-2,因此只要证-(x1+x2)-(x1+x2),即证,因此只要证,由于x1,x2 R时,0,0,因此由基本不等式知显然成立,故原结论成立.13.(1)解设等差数列 an的公差为d.因为

27、所以解得(q=-4 舍去).故an=3+3(n-1)=3n,bn=3n-1.(2)证明因为Sn=,所以所以+因为n1,所以 0,所以1-1,所以所以+|精.|品.|可.|编.|辑.|学.|习.|资.|料.*|*|*|*|欢.|迎.|下.|载.第 5 页，共 5 页文档编码:CV2L6B4B7Q8 HY7H8O7H2S6 ZU3L1S7E7E6文档编码:CV2L6B4B7Q8 HY7H8O7H2S6 ZU3L1S7E7E6文档编码:CV2L6B4B7Q8 HY7H8O7H2S6 ZU3L1S7E7E6文档编码:CV2L6B4B7Q8 HY7H8O7H2S6 ZU3L1S7E7E6文档编码:CV2

28、L6B4B7Q8 HY7H8O7H2S6 ZU3L1S7E7E6文档编码:CV2L6B4B7Q8 HY7H8O7H2S6 ZU3L1S7E7E6文档编码:CV2L6B4B7Q8 HY7H8O7H2S6 ZU3L1S7E7E6文档编码:CV2L6B4B7Q8 HY7H8O7H2S6 ZU3L1S7E7E6文档编码:CV2L6B4B7Q8 HY7H8O7H2S6 ZU3L1S7E7E6文档编码:CV2L6B4B7Q8 HY7H8O7H2S6 ZU3L1S7E7E6文档编码:CV2L6B4B7Q8 HY7H8O7H2S6 ZU3L1S7E7E6文档编码:CV2L6B4B7Q8 HY7H8O7H2S6

29、 ZU3L1S7E7E6文档编码:CV2L6B4B7Q8 HY7H8O7H2S6 ZU3L1S7E7E6文档编码:CV2L6B4B7Q8 HY7H8O7H2S6 ZU3L1S7E7E6文档编码:CV2L6B4B7Q8 HY7H8O7H2S6 ZU3L1S7E7E6文档编码:CV2L6B4B7Q8 HY7H8O7H2S6 ZU3L1S7E7E6文档编码:CV2L6B4B7Q8 HY7H8O7H2S6 ZU3L1S7E7E6文档编码:CV2L6B4B7Q8 HY7H8O7H2S6 ZU3L1S7E7E6文档编码:CV2L6B4B7Q8 HY7H8O7H2S6 ZU3L1S7E7E6文档编码:CV2

30、L6B4B7Q8 HY7H8O7H2S6 ZU3L1S7E7E6文档编码:CV2L6B4B7Q8 HY7H8O7H2S6 ZU3L1S7E7E6文档编码:CV2L6B4B7Q8 HY7H8O7H2S6 ZU3L1S7E7E6文档编码:CV2L6B4B7Q8 HY7H8O7H2S6 ZU3L1S7E7E6文档编码:CV2L6B4B7Q8 HY7H8O7H2S6 ZU3L1S7E7E6文档编码:CV2L6B4B7Q8 HY7H8O7H2S6 ZU3L1S7E7E6文档编码:CV2L6B4B7Q8 HY7H8O7H2S6 ZU3L1S7E7E6文档编码:CV2L6B4B7Q8 HY7H8O7H2S6

31、 ZU3L1S7E7E6文档编码:CV2L6B4B7Q8 HY7H8O7H2S6 ZU3L1S7E7E6文档编码:CV2L6B4B7Q8 HY7H8O7H2S6 ZU3L1S7E7E6文档编码:CV2L6B4B7Q8 HY7H8O7H2S6 ZU3L1S7E7E6文档编码:CV2L6B4B7Q8 HY7H8O7H2S6 ZU3L1S7E7E6文档编码:CV2L6B4B7Q8 HY7H8O7H2S6 ZU3L1S7E7E6文档编码:CV2L6B4B7Q8 HY7H8O7H2S6 ZU3L1S7E7E6文档编码:CV2L6B4B7Q8 HY7H8O7H2S6 ZU3L1S7E7E6文档编码:CV2

32、L6B4B7Q8 HY7H8O7H2S6 ZU3L1S7E7E6文档编码:CV2L6B4B7Q8 HY7H8O7H2S6 ZU3L1S7E7E6文档编码:CV2L6B4B7Q8 HY7H8O7H2S6 ZU3L1S7E7E6文档编码:CV2L6B4B7Q8 HY7H8O7H2S6 ZU3L1S7E7E6文档编码:CV2L6B4B7Q8 HY7H8O7H2S6 ZU3L1S7E7E6文档编码:CV2L6B4B7Q8 HY7H8O7H2S6 ZU3L1S7E7E6文档编码:CV2L6B4B7Q8 HY7H8O7H2S6 ZU3L1S7E7E6文档编码:CV2L6B4B7Q8 HY7H8O7H2S6 ZU3L1S7E7E6文档编码:CV2L6B4B7Q8 HY7H8O7H2S6 ZU3L1S7E7E6文档编码:CV2L6B4B7Q8 HY7H8O7H2S6 ZU3L1S7E7E6文档编码:CV2L6B4B7Q8 HY7H8O7H2S6 ZU3L1S7E7E6文档编码:CV2L6B4B7Q8 HY7H8O7H2S6 ZU3L1S7E7E6文档编码:CV2L6B4B7Q8 HY7H8O7H2S6 ZU3L1S7E7E6文档编码:CV2L6B4B7Q8 HY7H8O7H2S6 ZU3L1S7E7E6

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关键词：: 2021 福建专用高考数学复习第七不等式推理证明课时规范 35 直接间接新人

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本文标题：2021年福建专用高考数学总复习第七章不等式推理与证明课时规范练35直接证明与间接证明理新人教A版.pdf
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