求解两道高一数学题
\u6c42\u89e3\u4e24\u9053\u9ad8\u4e00\u6570\u5b66\u9898\u4e00\u89e3\uff1a(1)\u2235\u5728\u6570\u5217{a[n]}\u4e2d\uff0ca[n+1]=(1+1/n)a[n]+(n+1)/2^n
\u2234\u4e24\u8fb9\u9664\u4ee5(n+1)\uff0c\u5e76\u52a0\u4e0a2/2^(n+1)\uff0c\u5f97\uff1a
a[n+1]/(n+1)+2/2^(n+1)=a[n]/n+2/2^n
\u2235a[1]=1
\u2234{a[n]/n+2/2^n}\u662f\u5e38\u6570\u4e3aa[1]/1+2/2^1=2\u7684\u5e38\u6570\u6570\u5217
\u5373\uff1aa[n]/n+2/2^n=2
\u2234a[n]/n=2-2/2^n=2(1-1/2^n)
\u2235b[n]=a[n]/n
\u2234b[n]=2(1-1/2^n)
(2)\u2235a[n]=2(n-n/2^n)
\u2234S[n]=2[(1+2+3+...+n)-(1/2^1+2/2^2+3/2^3+...+n/2^n)]
\u4ee4\uff1aT[n]=1/2^1+2/2^2+3/2^3+...+n/2^n
\u5219\uff1aT[n]/2=1/2^2+2/2^3+3/2^4+...+n/2^(n+1)
\u5c06\u4e0a\u9762\u4e24\u5f0f\u76f8\u51cf\uff0c\u5f97\uff1a
T[n]/2=(1/2^1+1/2^2+1/2^3+...+1/2^n)-n/2^(n+1)
=(1/2)[1-(1/2)^n]/(1-1/2)-n/2^(n+1)
=1-(1/2)^n-n/2^(n+1)
=1-(n+2)(1/2)^(n+1)
\u2234T[n]=2-(n+2)(1/2)^n
\u2234S[n]=2{n(n+1)/2-[2-(n+2)(1/2)^n]}
=n(n+1)+(n+2)(1/2)^(n-1)-4
\u4e8c\u89e3\uff1a(1)\u2235\u6570\u5217{2a[n+1]-a[n]}\u662f\u516c\u6bd4\u4e3a1/3\u7684\u7b49\u6bd4\u6570\u5217
\u2234(2a[n+1]-a[n])/(2a[n]-a[n-1])=1/3
\u6574\u7406\uff0c\u5f97\uff1a6a[n+1]=5a[n]-a[n-1]
\u5373\uff1a2a[n+1]-a[n]=(2a[n]-a[n-1])/3 \u30101\u3011
\u2235a[1]=1/3\uff0ca[2]=5/18
\u2234{2a[n+1]-a[n]}\u662f\u9996\u9879\u4e3a2a[2]-a[1]=2/9\uff0c\u516c\u6bd4\u4e3a1/3\u7684\u7b49\u6bd4\u6570\u5217
\u5373\uff1a2a[n+1]-a[n]=(2/9)(1/3)^(n-1)=2(1/3)^(n+1)
\u2234a[n+1]+2(1/3)^(n+1)=[a[n]+2(1/3)^n]/2 \u30102\u3011
\u2234{a[n]+2(1/3)^n}\u662f\u9996\u9879\u4e3aa[1]+2(1/3)^1=1\uff0c\u516c\u6bd4\u4e3a1/2\u7684\u7b49\u6bd4\u6570\u5217
\u5373\uff1aa[n]+2(1/3)^n=(1/2)^(n-1)
\u2234a[n]=(1/2)^(n-1)-2(1/3)^n
(2)S[n]
=[1-(1/2)^n]/(1-1/2)-2(1/3)[1-(1/3)^n]/(1-1/3)
=2[1-(1/2)^n]-[1-(1/3)^n]
=2-(1/2)^(n-1)-1+(1/3)^n
=1-(1/2)^(n-1)+(1/3)^n
\u3010\u6b64\u9898\u7ed9\u51fa\u7684\u4e24\u4e2a\u6761\u4ef6\uff1a{Log[2](3a[n+1]-a[n])}\u662f\u516c\u5dee\u4e3a-1\u7684\u7b49\u5dee\u6570\u5217\uff0c\u4ee5\u53ca\u6570\u5217{2a[n+1]-a[n]}\u662f\u516c\u6bd4\u4e3a1/3\u7684\u7b49\u6bd4\u6570\u5217\uff0c\u5176\u5b9e\u662f\u7b49\u4ef7\u7684\u3002 \u30101\u3011\u3001\u30102\u3011\u5f0f\u53ef\u4ee5\u7528\u5f85\u5b9a\u7cfb\u6570\u6cd5\u83b7\u5f97\uff0c\u4f46\u7531\u4e8e\u90fd\u6bd4\u8f83\u7b80\u5355\uff0c\u76f4\u63a5\u89c2\u5bdf\u5373\u53ef\u83b7\u5f97\u3002\u3011
\u89e3\uff1a\u4f9d\u9898\u610f\uff0c\u51fd\u6570g\uff08x\uff09=loga\uff08a2x+t\uff09\uff08a\uff1e0\uff0ca\u22601\uff09\u5728\u5b9a\u4e49\u57df\u4e0a\u4e3a\u5355\u8c03\u9012\u589e\u51fd\u6570\uff0c\u4e14t\u22650\uff0c
\u800ct=0\u65f6\uff0cg\uff08x\uff09=2x\u4e0d\u6ee1\u8db3\u6761\u4ef6\u2461\uff0c
\u2234t\uff1e0\uff0e\u8bbe\u5b58\u5728[m\uff0cn]\uff0c\u4f7f\u5f97g\uff08x\uff09\u5728[m\uff0cn]\u4e0a\u7684\u503c\u57df\u4e3a[m\uff0cn]\uff0c
\u2234 \uff0c\u5373 \uff0c
\u2234m\uff0cn\u662f\u65b9\u7a0b\uff08ax\uff092-ax+t=0\u7684\u4e24\u4e2a\u4e0d\u7b49\u5b9e\u6839\uff0c
\u2234\u25b3=1-4t\uff1e0\uff0c
\u2234 \uff0c
\u6545\u9009D\uff0e
2f(2)-3f(1)=5 代入转换为 2(2a+b)-3(a+b)=5
2f(0)-f(-1)=2 代入转换为 2b-(-a+b)=2
上述两式结合 得b=-3/2 a=7/2
所以f(x)=7/2 x-3/2
2.已知f(2x+1)=x^2-2x 设 t=2x+1 则 x=(t-1 )/2
f(t)=[(t-1 )/2]^2-2[(t-1 )/2]=[(t^2-2t+1)/4]-t+1
将t转换为x f(x)=[(x^2-2x+1)/4]-x+1
√ 此符号为平方根 下同
f(√2) 2x+1=√2 x=（√2-1）/2 将x的得值代入f(2x+1)=x^2-2x
得：（3-√2）/4-√2+1 所以f(√2)=（3-√2）/4-√2+1
问题呢？？？
绛旓細瑙ｏ細锛1锛夌敱棰樻剰鍙煡褰撹鍒惰嵂鍘傛瘡澶╁簾姘斿鐞嗛噺璁″垝涓20鍚ㄦ椂锛 姣忓ぉ鍒╃敤璁惧澶勭悊搴熸皵鐨勭患鍚堟垚鏈负(20)402012002000f鍏冿紝鈥︹︹2鍒 杞寲鐨勬煇绉嶅寲宸ヤ骇鍝佸彲寰楀埄娑80201600鍏冿紝 鈥︹︹3鍒 鎵浠ュ伐鍘傛瘡澶╅渶瑕佹姇鍏ュ簾姘斿鐞嗚祫閲戜负400鍏冿紟 鈥︹...
绛旓細1.鈻斥墺0 (4a)^2-4*1*(-4a+3)鈮0 (a-1)^2-4*1*a^2鈮0 (2a)^2-4*1*(-2a)鈮0 鍒嗗埆姹傚嚭a鐨勮寖鍥村彇骞堕泦 鏈缁堢瓟妗堝簲涓猴細a<=-3/2鎴朼>=-1 2.A鈭〣鈮犵┖闆 -2鈭圓 x^2+px+q=0 (1)qx^2+px+1=0 (2)涓や釜鏂圭▼鏈変竴涓浉鍚岀殑瀹炴暟鏍 涓 -2鏄柟绋(1)鐨勮В 鐩稿悓鐨勬牴涓簒...
绛旓細1.C 2.f(3+x)=f(3-x),鍒欙細f(x)=f(6-x)鎵浠ワ紝濡俛鏄痜(x)=0鐨勬牴锛屽垯锛6-a涔熸槸鏍 涓ゆ牴涔嬪拰=6锛6涓笉鍚岀殑瀹炴暟鏍瑰彲浠ュ垎鎴3缁勶紝鎵浠6涓疄鏁版牴鐨勭殑鍜屼负锛3*6=18 绛旀锛欴 3.璁緔=log2(x),鍒欙細x=2^y 鎵浠ワ細f(y)=f(log2(x))=(x-1)/(x+1)=(2^y-1)/(2^y+1)f...
绛旓細1 瑙ｏ細16sin6掳sin42掳sin66掳sin78掳=16sin6掳cos48掳cos24掳cos12掳 =16sin6掳cos6掳cos12掳cos24掳cos48掳/cos6掳 =8sin12掳cos12掳cos24掳cos48掳/cos6掳 =4sin24掳cos24掳cos48掳/cos6掳 =2sin48掳cos48掳/cos6掳 =sin96掳/cos6掳 =cos6掳/cos6掳 =1 2瑙ｏ細f(x)=...
绛旓細涓 鑻鏄疊鐨勭湡瀛愰泦 鍒欐湁 a>5鎴朼+4鈮-1 瑙ｄ笉绛夊紡寰 a>5鎴朼鈮-5 浜 P涓 x=m^2+3m+1=m^2+3m+9/4-9/4+1=(m+3/2)^2-9/4+1=(m+3/2)^2-5/4鈮-5/4 鍚岀悊 T涓 x=(n-3/2)^2-5/4鈮-5/4 鎵浠鈭猅=P鈭㏕鈮-5/4 鈶ｆ纭 ...
绛旓細绗浜岄锛氬師绛夊紡鍙寲涓9(x-1/3)^2+6y^2=1 璁緓=1/3+1/3cosa y=1/鏍瑰彿6*sina x^2+y^2=5/18-(1/18*(cosa)^2-2/9cosa)褰揷osa=1/2鏃 鍙栨渶澶т负3/8
绛旓細鍥炵瓟锛氱涓棰13
绛旓細1.瀵圭О杞存槸x=5/2,鍑芥暟鍦ㄣ-1锛6銆戜笂鍏堝噺鍚庡锛寉min=f(5/2)=鈭11/2 ymax绛変簬涓や釜绔偣涓殑鏈澶х殑涓涓紝f(-1)=f(6)=鈭15 鎵浠MAX=鈭15 鍊煎煙涓恒愨垰11/2,鈭15銆2.閫嗘眰娉曪細涔熷氨鏄粠寮忎簡涓В鍑簒^4,鍐嶅埄鐢▁^4鐨勬湁鐣屾ф帹鍑簓鐨勮寖鍥 1-x^4=y+2yx^4 x^4=(1-y)/(2y-1)...
绛旓細涓涓厓绱狅紝璇存槑螖=0锛埼=b^2-4*ac) 螖=64-64k=0,寰梜=1, 鎵浠 x=4 螖=9-4a 褰揳=0鏃,x=1/3 褰0<a<9/4鎴朼<0鏃讹紝螖>0锛寈=(3+鈭(9-4a))/2a 鎴杧=(3-鈭(9-4a))/2a 锛堝叕寮忔硶锛夊綋a>9/4鏃讹紝螖<0锛屾槸绌洪泦 褰揳=9/4鏃讹紝x=2/3 鍐嶇敤闆嗗悎鐨勫舰寮忚〃绀哄氨濂戒簡 ...
绛旓細1銆佸綋x>=0鏃讹紝f(x)=3x,鍗曡皟澧烇紱褰搙<=0鏃讹紝f(x)=-3x,鍗曡皟鍑忥紱2銆佸崟璋冨锛岃x1>x2 f(x1)-f(x2)=(x1)³-(x2)³+(x1-x2)=(x1-x2)(x1²+x2²+x1*x2)+(x1-x2)=(x1-x2)(x1²+x2²+x1*x2+1)=(x1-x2)[(x1+0.5x2锛²...
