{高一数学}对数的运算,请教几道题。在线等!!!!!!!!!!~~~~~~ 高一数学对数的运算 化简2道题

\u5e2e\u5fd9\u89e3\u51e0\u4e2a\u9ad8\u4e00\u51fd\u6570\u6570\u5b66\u9898\uff08\u9ad8\u5206\uff0c\u5728\u7ebf\u7b49~\uff01\uff01\uff01\uff09

1. x-1>0\uff0c
3-x>0
\u52191<x<3.

(2+b)/2<x<b+1,
B\u662fA\u7684\u5b50\u96c6\uff0c\u5f53B\u4e3a\u7a7a\u96c6
b+1\u2264(2+b)/2,\u5373b\u22640
B\u4e0d\u4e3a\u7a7a\u96c6\u65f6\uff0c
b>0\uff0c\u4e14b+1\u22643\u4e141\u2264(2+b)/2\uff0c
\u5f97\uff1a0<b\u22642
\u7efc\u5408\u5f97\uff1ab\u22642

2.\u51fd\u6570\u5f00\u53e3\u5411\u4e0a\uff0c\u4e14\u8fc7\uff080\uff0c2\uff09\u70b9\uff0c
\u6240\u4ee5\u6ee1\u8db3\u5df2\u77e5\u6761\u4ef6\u5e94\u8be5\u662f\uff1a\u25b3>0,\u5bf9\u79f0\u8f74\u5728y\u8f74\u53f3\u4fa7\uff1a-b/2a>0,\u89e3\u5f97\uff1aa<-\u221a2
\uff082\uff09f(x+1)=f(1-x),\u5373\u5bf9\u79f0\u8f74\u662fx=1\uff0c\u6240\u4ee5a=-1\uff0c\u5f00\u53e3\u5411\u4e0a\uff0c
\u5728x=-5\u5904\u53d6\u5f97\u6700\u5927\u503c\uff1a37\uff1b\u5728x=1\u5904\u53d6\u6700\u5c0f\u503c\uff1a1
\uff083\uff09\u8981\u8ba8\u8bba\u5bf9\u79f0\u8f74\u548c\u533a\u95f4\u7684\u5173\u7cfb\uff0c\u5bf9\u79f0\u8f74\u4e3a\uff1ax=-a,\u5f00\u53e3\u5411\u4e0a\u3002
-a\u2264-5\uff0c\u5373a\u22655\u65f6\uff0c\u5bf9\u79f0\u8f74\u5728\u533a\u95f4\u7684\u5de6\u4fa7\uff0c\u2234\u5728x=-5\u65f6\u53d6\u5f97\u6700\u5c0f\u503c\uff1a27-10a
-5<-a<5,\u5373-5<a<5\u65f6\uff0c\u5bf9\u79f0\u8f74\u5728\u533a\u95f4\u5185\uff0c\u5728x=-a\u65f6\u53d6\u5f97\u6700\u5c0f\u503c\uff1a2-a²
-a\u22655\uff0c\u5373a\u2264-5\u65f6\uff0c\u5bf9\u79f0\u8f74\u5728\u533a\u95f4\u53f3\u4fa7\uff0c\u5728x=5\u65f6\u53d6\u5f97\u6700\u5c0f\u503c\uff1a27+10a

3.
h(-x)=loga (1-x)-loga (1+x)=-h(x),
\u5219h(x)\u4e3a\u5947\u51fd\u6570

f(3)=2,
\u5219loga4=2
\u5219a=2
h(x)>0\uff0c\u5373loga (1+x)-loga (1-x)=loga(1+x)/(1-x)>0
\u800ca>1,
\u5219(1+x)/(1-x)>1
\u4e14\u8981\u4f7f\u5f97\u5bf9\u6570\u51fd\u6570\u6709\u610f\u4e49\uff0c\u5219
1+x>0
1-x>0
\u52190<x<1



4.y=g(mx^2+2x+m)\u7684\u503c\u57df\u4e3aR,\u5373
log1/3 (mx^2+2x+m) \u503c\u57df\u4e3aR
\u53ef\u77e5\u6709mx^2+2x+m>=0
\u5f53m\u22600\u65f6
\u6240\u4ee5\u25b3>=0
\u52194-4m^2>=0
\u5219-1>=m>=1
\u5f53m=0\u65f6\uff0c\u4ecd\u7136\u6210\u7acb
\u51fd\u6570y=[f(x)]^2-2af(x)+3\u7684\u6700\u5c0f\u503ch(a)
y=[f(x)-a]^2+3-a^2
\u5f53a<1/3\uff0c\u5219\u53ef\u77e5\u5728x=1\u65f6\u3002\u53d6\u5f97\u6700\u5c0f\u503c:4-2a
\u5f531/3<=a<=3\uff0c\u5219\u53ef\u77e5\u5728x=log1/3 a\u65f6\u3002\u53d6\u5f97\u6700\u5c0f\u503c3-a^2
\u5f53a>3\uff0c\u5219\u53ef\u77e5\u5728x=-1\u65f6\u3002\u53d6\u5f97\u6700\u5c0f\u503c:4+2a

\u51fd\u6570y=g[f(x^2)]\u7684\u5b9a\u4e49\u57df\u4e3a[m,n],\u503c\u57df\u4e3a[2m,2n],
\u5373y=x^2
\u5219\u53ef\u80fd\u4e3a\uff1a
m>=0
\u5219\uff0c\u53ea\u80fd\u4e3a\uff1a
2m=m^2
2n=n^2
\u5219m=0\uff0cn=2

\u6216\u8005
n<0
\u5219\uff0c\u53ea\u80fd\u4e3a\uff1a
2m=n^2
2n=m^2
\u65e0\u89e3

m0
\u5219\u3002
\u6700\u5c0f\u503c\u4e3ax=0 ,\u53732m=0, m=0
\u4e0d\u6ee1\u8db3\uff0c\u820d\u53bb









y=log1/2 (x^2-2x+5)\u7684\u503c\u57df

x^2-2x+5=(x-1)^2+4
\u800c(x-1)^2+4 \u5f53x=1\u53d6\u5f97\u6700\u5c0f\u503c4
\u800cy=log1/2 (x^2-2x+5)\uff0c\u6307\u65701/2<1,\u51fd\u6570\u5355\u51cf\u3002
\u6240\u4ee5y=log1/2 (x^2-2x+5)\u6709\u6700\u5927\u503c
y=log1/2 4=-2
\u6240\u4ee5\u503c\u57df\u4e3a[-\u221e\uff0c-2]


\u5e26\u51651\uff0c\u33d11+2-3=-1<0
\u5e26\u51652\uff0c\u33d12+4-3=l.3>0
\u5219\u6839\u5fc5\u5728B(1,2)

1.\u4e3a\u4e86\u6253\u5b57\u65b9\u4fbf\uff0c\u6211 \u5148\u8bbea=lg5,b=lg2
\u90a3\u4e48\u6709a+b=lg(5*2)=1
\u6240\u4ee5\u539f\u5f0f=a³+3ab+b³
=(a+b)(a²-ab+b²)+3ab
=1*\uff08a²-ab+b²)+3ab
=(a+b)²
=1
2.\u8fd9\u91cc\u6253\u4e0d\u4e0a\u5e95\u6570\uff0c\u6545\u7565\u53bb\u4e86\uff0c\u4f60\u5199\u7684\u65f6\u5019\u518d\u8865\u4e0a\u5427\u3002O(\u2229_\u2229)O
\u7b2c\u4e00\u4e2a\u5bf9\u6570=1/2log7-1/2log48
\u7b2c\u4e8c\u4e2a\u5bf9\u6570=1/2log144
\u7b2c\u4e09\u4e2a\u5bf9\u6570\u4e0d\u53d8
\u90a3\u4e48\u539f\u5f0f=1/2\uff08log7-log48+log144-log42\uff09
=1/2log[(7*144)/(48*42)]
=1/2log(2^-1)
=-1/2long2
=-1/2

解:(1)原式=[lg3+(2/5)(2lg3)+(3/5)(3lg3/2)-lg3/2]/(4lg3-3lg3) (应用对数换底公式)
=(11/5)lg3/(lg3)
=11/5;
(2)原式=[(2lg3)/(3lg2)]*[(5lg2)/(lg3)] (应用对数换底公式)
=10/3;
(3)原式=(lg5+lg7)/lg5+lg2/(-lg2)+(2lg5+lg2)/lg5-(lg7+lg2)/lg5 (应用对数换底公式)
=3lg5/lg5-1
=3-1
=2;
(4)原式=(2lg3)/(lg3/2)+(3lg3)/(2lg3)+(1/4)^[(-4lg2)/(2lg2)] (应用对数换底公式)
=4+3/2+4²
=43/2。

过程在图片中,细节自己推一下



  • {楂樹竴鏁板}瀵规暟鐨勮繍绠,璇锋暀鍑閬撻銆傚湪绾跨瓑!!!~~~
    绛旓細瑙o細锛1锛夊師寮=[lg3+(2/5)(2lg3)+(3/5)(3lg3/2)-lg3/2]/(4lg3-3lg3) (搴旂敤瀵规暟鎹㈠簳鍏紡)=(11/5)lg3/(lg3)=11/5锛涳紙2锛夊師寮=[(2lg3)/(3lg2)]*[(5lg2)/(lg3)] (搴旂敤瀵规暟鎹㈠簳鍏紡)=10/3锛涳紙3锛夊師寮=(lg5+lg7)/lg5+lg2/(-lg2)+(2lg5+lg2)/lg5-...
  • 瀵规暟鏄珮鍑犵殑鍐呭鍟娿
    绛旓細瀵规暟鏄楂樹竴鏁板蹇呬慨涓瀛︾殑銆瀵规暟鐨勮繍绠娉曞垯锛1銆乴og(a) (M路N锛=log(a) M+log(a) N 2銆乴og(a) (M梅N)=log(a) M-log(a) N 3銆乴og(a) M^n=nlog(a) M 4銆乴og(a)b*log(b)a=1 5銆乴og(a) b=log (c) b梅log (c) a 瀵规暟搴旂敤 瀵规暟鍦ㄦ暟瀛﹀唴澶栨湁璁稿搴旂敤銆傝繖浜涗簨浠朵腑鐨...
  • 楂樹竴鏁板瀵规暟鐨勮繍绠鍏紡 鍙婅瑙
    绛旓細鐢辨崲搴曞叕寮 log(a)(b)=log(b)(b)/log(b)(a) ---鍙栦互b涓哄簳鐨瀵规暟,log(b)(b)=1 =1/log(b)(a)杩樺彲鍙樺舰寰:log(a)(b)*log(b)(a)=1
  • 姹傞珮涓鏁板蹇呬慨涓鎸囨暟瀵规暟鐨勮绠鍏紡
    绛旓細瀵规暟鐨勮繍绠娉曞垯锛1銆乴og(a) (M路N锛=log(a) M+log(a) N 2銆乴og(a) (M梅N)=log(a) M-log(a) N 3銆乴og(a) M^n=nlog(a) M 4銆乴og(a)b*log(b)a=1 5銆乴og(a) b=log (c) b梅log (c) a 鎸囨暟鐨勮繍绠楁硶鍒欙細1銆乕a^m]脳[a^n]=a^(m锛媙) 銆愬悓搴曟暟骞傜浉涔,搴曟暟涓...
  • 楂樹腑鏁板瀵规暟鍏紡澶у叏
    绛旓細楂樹腑鏁板瀵规暟鍏紡澶у叏濡備笅锛1銆瀵规暟杩愮畻娉曞垯锛歛^log锛坅锛塏=N锛坅>0涓攁涓嶇瓑浜1)锛塴og锛坅锛塣n=n锛坅>0涓攁涓嶇瓑浜1锛塴og锛坅锛塎N=log锛坅锛塎+log锛坅锛塏锛坅>0鏈坅涓嶇瓑浜1锛夈俵og锛坅锛塎/N=log锛坅锛塎-log锛坅锛塏锛坅>0鏈坅涓嶇瓑浜1锛夈俵og锛坅锛塣M^n=nlog锛坅锛塣M锛坅>0鏈坅涓嶇瓑浜1锛夈...
  • 瀵规暟鍜屾寚鏁鐨勮繍绠鍏紡鍒嗗埆鏄粈涔?
    绛旓細瀵规暟鐨勮繍绠鍏紡锛1銆乴og(a) (M路N锛=log(a) M+log(a) N 2銆乴og(a) (M梅N)=log(a) M-log(a) N 3銆乴og(a) M^n=nlog(a) M 4銆乴og(a)b*log(b)a=1 5銆乴og(a) b=log (c) b梅log (c) a 鎸囨暟鐨勮繍绠楀叕寮忥細1銆乕a^m]脳[a^n]=a^(m锛媙) 銆愬悓搴曟暟骞傜浉涔,搴曟暟涓...
  • 楂樹竴瀵规暟鍏紡
    绛旓細楂樹竴瀵规暟鍏紡鐨勫洖绛斿涓:瀵规暟鏄鏁板涓竴涓噸瑕佽屽箍娉涘簲鐢ㄧ殑姒傚康锛屽鏁板叕寮忋佸畾涔夈佺敤娉曠瓑鏂归潰鐨勪簡瑙o紝鏈夊姪浜庡湪瑙e喅鍚勭闂涓洿鐏垫椿鍦拌繍鐢ㄦ暟瀛︾煡璇嗐備互涓嬫槸瀵瑰鏁扮浉鍏冲唴瀹圭殑璇︾粏浠嬬粛銆瀵规暟鐨鍩烘湰瀹氫箟锛氬鏁版槸骞杩愮畻鐨勯嗚繍绠椼傚浜庢瀹炴暟a銆乥鍜屾鏁存暟n锛屽鏋渁=b锛屽垯璁颁綔n=logb锛屽叾涓璦绉颁负搴曟暟锛宐绉颁负...
  • 楂樹竴鏁板,姹傚ぇ绁,瀵规暟杩愮畻
    绛旓細鈶㈠彸杈=log2[4/(x+1)]=宸﹁竟=log2(x-1)锛屽緱 4/(x+1)=(x-1)锛0 x=鏍瑰彿5 鈶e師鏂圭▼鍙寲涓 x^2-ax-3a=0 鍦(3锛4)鏈夎В锛屽緱(9-3a-3a)(16-4a-3a)锛0 a鈭(3/2,16/7)鈶ュ師寮=3og3(2)/2 * 5log2(3)/6=5/4 鈶﹀師寮=1/2+3/2+13=15 鈶ц瘉鏄 [1+(...
  • 姹(楂樹竴鏁板)瀵规暟杩愮畻鐨鍏紡
    绛旓細2銆乴og(a)(a^b)=b 3銆乴og(a)(MN)=log(a)(M)+log(a)(N);4銆乴og(a)(M梅N)=log(a)(M)-log(a)(N);5銆乴og(a)(M^n)=nlog(a)(M)6銆乴og(a^n)M=1/nlog(a)(M)鍙傝冭祫鏂欙細http://zhidao.baidu.com/question/170168940.html?an=0&si=2 ...
  • 楂樹腑鏁板,瀵规暟鐨勮繍绠
    绛旓細璇︾粏姝ラ鍐欏湪绾镐笂浜嗭紝琛屽姝hВ
  • 扩展阅读:高一数学对数讲解视频 ... 高一下数学公式大全 ... 高一数学对数公式大全 ... 高一数学对数函数视频 ... 高一数学上册所有公式 ... 高一数学必背公式 ... 高一数学例题大全 ... 高一数学思维导图 ... 高一数学对数运算教学视频 ...

    本站交流只代表网友个人观点,与本站立场无关
    欢迎反馈与建议,请联系电邮
    2024© 车视网