请教高一数学题目
\u8bf7\u6559\u9ad8\u4e00\u6570\u5b66\u9898Sn=4a(n-1)+2
S(n+1)-Sn=4an-4a(n-1)=a(n+1)
[a\uff08n+1\uff09-2an]/[an-2a(n-1)]=[2an-4a(n-1)]/[an-2a(n-1)]=2
\u8bbebn=a\uff08n+1\uff09-2an
b1=3
bn=3\u00d72^(n-1)
\u6240\u4ee5a\uff08n+1\uff09-2an=3\u00d72^(n-1)
\u4e0a\u5f0f\u53ef\u5316\u4e3aa\uff08n+1\uff09-3\uff08n+1)2^(n-1)=2[an-3n*2^(n-2)]
[a\uff08n+1\uff09-3\uff08n+1)2^(n-1)]/[an-3n*2^(n-2)]=2
\u4ee4cn=an-3n*2^(n-2)
\u5219c1=-1/2
cn=-2^\uff08n-2)
\u5219an=3n*2^(n-2\uff09-2^\uff08n-2)
=\uff083n-1\uff092^(n-2)
\u7531\u9898\u610f\u77e5\uff0c\u4e09\u89d2\u5f62ABD\u662f\u76f4\u89d2\u4e09\u89d2\u5f62\u3002\u6240\u4ee5AD=\u6839\u53f73\uff0cAB=1\uff0c
\u4e09\u89d2\u5f62ADC\u4e2d\uff0c\u7531\u4f59\u5f26\u5b9a\u7406\uff0ccos150\u00b0=(1+3-AC²)/2\u6839\u53f73 \u6240\u4ee5AC=\u6839\u53f77
ADC\u9762\u79ef=1/2*AD*DC*sin150=\u6839\u53f73/4
ABD\u9762\u79ef=\u6839\u53f73/2
\u6240\u4ee5ABC\u9762\u79ef=\u6839\u53f73/6
明显是一次函数，y=8*x，八千克为64
y=40/5 *x=8xy=8*8=64
绛旓細3x(x+5)锛3x2+7 x-4 < 2x+1 3x+14 > 4(2x-9)3x-7鈮4x-4 2x-3x-3锛6 0.4(x-1)鈮0.3-0.9x x-4 < 2x+1 2x-6 < x-2 3脳10x<500 7锛圶+3锛>98 2x-3x+3锛6 2x-3x+1锛6 2x-3x+3锛1 2x-19锛7x+31 3x-2(9-x)锛3(7+2x)-(11-6x)2(3x-1)-3(4x+5...
绛旓細1銆佸浠绘剰瀹炴暟x,y鏈塮(x+y)=f(x)*f(y). 浠=0銆俧(0)涓嶇瓑浜0锛屼护y=0銆傚垯鏈塮(x+0)=f(x)*f(0)f(x)=f(x)*f(0)寰梖(0)=1 鎵浠(0)=f(x-x)=f(x)*f(-x)=1 x>0鏃躲俧(x)>1 鎵浠-x<0.f(-x)=1/f(x) 鎵浠0<f(-x)<1 鍗硏<0.0<f(x)<1寰楄瘉 2銆佽...
绛旓細鏁咃細b=0锛宎=-1 鏁咃細a^2008+b^2007锛1
绛旓細a1+a4=a1+a1*q^3=a1(1+q^3)=18 a2+a3=a1*q+a1*q^2=a1q(1+q)=12 涓ゅ紡鐩搁櫎锛屽緱锛(1-q+q^2)/q = 3/2 瑙ｅ緱锛歲=2鎴杚=1/2 鍥犱负锛屽叕姣攓鏄暣鏁 鎵浠ワ紝q=2 鍒欙細a1=2 S8=510 閫塂
绛旓細x>a鏃 f(x)=x^2+x-a+1=(x+1/2)^2-a+3/4 f(x)鏈灏忓间负-a+3/4 杩欐椂-a+3/4>a 鍗砤<3/8锛屽惁鍒欐棤瑙 x<a鏃 f(x)=x^2-x+a+1=(x-1/2)^2+a+3/4 f(x)鏈灏忓间负a+3/4 a+3/4<a 鏃犺В 鎵浠<3/8锛 f(x)鏈灏忓间负-a+3/4 ...
绛旓細z=log(6)k (1)瑙ｏ細瑕佷娇 2x=py p=2x/y 鍗 p=2log(3)k梅log(4)k 鍒檖=2*log(3)4 鎴栬呰p=log(3)16 (2)璇佹槑锛1/2y=1/2log(4)k=log(k)2 1/z=1/log(6)k=log(k)6 1/x=1/log(3)k=log(k)3 鎵浠1/z-1/x=log(k)6-log(k)3=log(k)2 璇佹槑瀹屾瘯銆傝繖棰樺叾瀹炲氨...
绛旓細瑙ｏ細锛1锛夊厛姹侫鈭狟锛氣埖 A = { x | 4 鈮 x < 8 },锛孊 =锝泋 | 2 < x < 10锝濓紝鈭 闆嗗悎A閲岀殑浠讳綍涓涓厓绱犲潎鍦ㄩ泦鍚圔涓傗埓 A鈭狟 = B =锝泋 | 2 < x < 10锝濄傚啀姹 A鐨勮ˉ闆 浜锛氣埖 A = { x | 4 鈮 x < 8 },锛屸埓 A鐨勮ˉ闆 = { x | x 锛 4 鎴 x...
绛旓細瑙ｏ細锛1锛塧=锛坰inx 1) b=锛坈osx,-0.5)鈭 a鈯 鈭碼路b=0 涓╝+b涓=鈭氾紙a锛媌锛²=鈭氾紙a²锛媌²锛2ab锛=鈭氾紙sin²x 锛1锛媍os²x锛1/4锛=鈭9/4=3/2 锛2锛塅(x)=a路锛坅-b锛=a²-ab =sin²x 锛1-锛1/2sin2x-1/2锛=锛1-cos&...
绛旓細cos(B)=2銆恈os(B/2)銆慯2-1=-3/5 寰 sin(B)=4/5 鐢棰樼洰鏈 sin(C)=cos(C)=(鏍2锛/2 鎵浠 sin(A)=sin(B+C)=sin(B)cos(C)+sin(C)cos(B)=(鏍2锛/10 鐢辨寮﹀畾鐞嗘湁 a/sin(A)=c/sin(C) 寰 c=10 鎵浠ラ潰绉疭=a*c*sin(B)/2=10*2*0.8/2=8 ...
绛旓細鍥炵瓟锛氳缁冪粌瀛椾簡銆傘傘
