项和，帮忙解答下高一数学题，，感谢！！！答案写清楚哦！！！！急急急急，速采纳的！！全部都做哦！ 19，帮忙解答下高一数学题，，感谢！！！答案写清楚哦！！！！...
\uff0c\u5e2e\u5fd9\u89e3\u7b54\u4e0b\u9ad8\u4e00\u6570\u5b66\u9898\uff0c\uff0c\u611f\u8c22\uff01\uff01\uff01\u7b54\u6848\u5199\u6e05\u695a\u54e6\uff01\uff01\uff01\u6025\u6025\u6025\u6025\uff0c\u901f\u91c7\u7eb3\u7684\uff01\uff01\u5168\u90e8\u90fd\u505a\u54e6\uff01\u56e0\u4e3a{an}\u4e3a\u7b49\u5dee\u6570\u5217a1=25\uff0ca4=16
\u6240\u4ee5d=(a4-a1)/3=-3
\u6240\u4ee5an=28-3n
\u3010\u6ce8\uff1a\u4ee428-3n=0\uff0cn=9.33••••••\uff0c\u7531\u6b64\u53ef\u786e\u5b9aan\u5927\u4e8e\u6216\u5c0f\u4e8e0\u57289\u548c10\u3011
\u6240\u4ee5a9=1\uff0ca10=-2
\u6240\u4ee5an<0\u662f\u4ece\u7b2c10\u9879\u5f00\u59cb\u7684
a1+a3+•••+a19=(a1+a19)\u00d710\u00f72=(25+28-3\u00d719)\u00d710\u00f72=-20
\u53ea\u8981\u662f\u7b49\u5dee\u6570\u5217\u6c42\u548c\u90fd\u7528\uff0c\u548c=(\u9996\u9879+\u5c3e\u9879)\u00d7\u9879\u6570\u00f72
\u7531\u9898\u610f\u53ef\u5f97\uff1aBC=50*1/2=25\uff1b\u2220BCA=25+80=105\uff1b\u2220CBA=30\uff1b\u5219\u2220A=45\uff1b
\u8bbeAC=b\uff1bBC=a=25\uff1bAB=c
\u9762\u79ef\u76f8\u7b49\uff0c\u5219\uff1a1/2*acsin30=1/2*bcsin45\uff1b\u6240\u4ee5\uff1aasin30=bsin45
b=asin30/sin45=25/2*2\u221a2/2\uff1d25\u221a2/2
\u6240\u4ee5\uff1a\u6b64\u65f6\u706f\u5854\u4e0eA\u7684\u8ddd\u79bb\u4e3a\uff1a25\u221a2/2 \u6d77\u91cc
∵等差数列前n项和公式为：Sn=na1+n(n-1)d/2
∴S10=10a1+10x(10-1)d/2，即10a1+45d=310……①
S20=20a1+20x(20-1)d/2，即20a1+190d=1220……②
联立两式解得：a1=4，d=6
则等差数列的前n项和为：Sn=4n+n(n-1)x6/2=3n²+n
2、等差数列的公差：d=(a20-a10)/10=2
首项为：a1=a10-9d=30-9x2=12
则Sn=na1+n(n-1)d/2
=12n+n(n-1)x2/2
=n²+11n
=242
即n²+11n-242=0
(n+22)(n-11)=0
解得：n=11或-22(舍去)
则n=11
【数学的快乐】团队为您解答！祝您学习进步
不明白可以追问！
满意请点击下面的【选为满意回答】按钮，O(∩_∩)O谢谢
数列是等差数列；所以设Sn=mn²+kn;
由S10=310；即：310=100m+10k; 化简得：k=31-10m;
由S20=1220=400m+20k; k=61-20m;
所以：31-10m=61-20m; m=3; k=1
所以Sn=3n²+n;
变式：
公差d=(a20-a10)/(20-10)=2; a10=a1+9d=a1+18; a1=12;
Sn=na1+n(n-1)×d/2=12n+n(n-1)=n²+11n=242;
n=11
an=a1+(n-1)d,Sn=n(a1+an)/2
S10=10(a1+a10)/2=310,
S20=20(a1+a20)/2=1220,
简化得a1+a10=62,
a1+a20=122,
a20-a10=60,
a1+19d-a1-9d=60,
d=6,
a1+a10=62,
a1+a1(10-1)d=62,
a1=4,
an=4+6(n-1),即：an=6n-2
Sn=na1+[n(n-1)d/2]
Sn=3n²+n
绛旓細鍒嗙被璁ㄨ锛屽綋n涓哄伓鏁帮紝姝ｅソ鏄痭涓2閰嶅瀹屾垚姹傚拰銆俷涓哄鏁帮紝澶氫簡涓涓礋鐨2n-1锛屾墍浠ユ槸n-1涓2姹傚拰鍔犱笂涓涓礋鐨2n-1
绛旓細C: (-x^1/4 y^-1/3)(x^1/4 y^2/3)(x^-1/2 y^2/3)=y璇ュ紡浣犲簲璇ュ鍐欎簡涓涓礋鍙凤紝搴旇鏄紙x^1/4 y^-1/3)(x^1/4 y^2/3)(x^-1/2 y^2/3)=y锛屽噯纭殑璇达紝杩欓」涔熸槸閿欒鐨 D:-锛15a^1/3 b^1/3 c^-3/4)梅锛25a^-1/3 b^1/3 c^5/4)=--3/5 a^2/...
绛旓細b²-b-2=0 (b-2)(b+1)=0 b=2 涓夎褰负绛夎叞涓夎褰紝a鍜宐绛夎叞 sinA=鏍瑰彿(1-cos²A)=(鏍瑰彿14)/4 sin2A=2sinAcosA=鏍瑰彿7/4 cos2A=3/4 2.cos(2A+蟺/3)=cos2Acos蟺/3-sin2Asin蟺/3=3/4*1/2-鏍瑰彿7/4*鏍瑰彿3/2=(3-鏍瑰彿21)/8 甯屾湜瀵逛綘鏈夋墍甯姪 濡傛湁闂锛...
绛旓細1锛堿n + An+1 - 1 = 2鍊 鏍瑰彿An x 鏍瑰彿 An+1 鍙宠竟绉诲乏杈 宸﹁竟1绉诲彸杈 鍦ㄨ繘琛岄厤鏂 灏辨湁 锛堟牴鍙蜂笅An+1锛 鈥 锛 鏍瑰彿涓 An 锛夋暣浣撶殑骞虫柟 = 1 鍙堢敱浜庢槸閫掑鏁板垪 鎵浠ユ湁 An = n骞虫柟 璇佹槑鐨勮瘽 鎴戞槸鐢ㄨ椤瑰拰鏀剧缉娉 寰呰瘉寮忎负 1+1/2²+1/3²+ ....
绛旓細绗竴涓細an=4n-2 锛 鍏48椤 绗簩涓細bn=6n-4 锛 鍏34椤 鍏叡鏁板垪锛歝n=12n-10 锛岀敱 12n-10 鈮 190 寰 n 鈮 16 鎵浠n=12路16路17/2-160 = 1472
绛旓細鎴戞潵瑙ｇ瓟鍚!!绗竴棰樼瓟妗堟槸:锛嶆牴鍙2锛滐紳a锛滐紳9/4 绗簩棰樼瓟妗堟槸:褰-2锛渁锛0鏃,涓嶇瓑寮忕殑瑙ｉ泦涓篟 褰揳=0鏃,涓嶇瓑寮忕殑瑙ｉ泦涓簕x|x涓嶇瓑浜1} 褰揳=-2鏃,涓嶇瓑寮忕殑瑙ｉ泦涓簕x|x涓嶇瓑浜-1} 褰揳锛0鎴栬卆锛-2鏃 涓嶇瓑寮忕殑瑙ｉ泦涓 锝泋锝渪锛瀉+1+鏍瑰彿(a^2+2a)鎴杧...
绛旓細瑙ｇ瓟濡備笅锛歛n涓轰簡濂藉尯鍒嗭紝涓嬮潰閮藉啓鎴怉n 锛1锛夈佸洜涓篠n=n锛圓n+1锛-n²鎵浠锛坣-1锛=锛坣-1锛塠A锛坣-1锛+1]-锛坣-1锛²涓婇潰涓や釜寮忓瓙浣滃樊鍙緱 An=Sn-S锛坣-1锛=[n锛圓n+1锛-n²]-锛坣-1锛塠A锛坣-1锛+1]+锛坣-1锛²绠鍖栧緱An-A(n-1)=2 鍒橝2-A1=2 A3-...
绛旓細1锛夎嫢a灞炰簬S锛岀敱棰樺彲鐭1/1-a灞炰簬S锛屾墍浠1/(1/1-a)=1-a灞炰簬S 2锛夎嫢2灞炰簬S锛岀敱棰1/1-2=-1锛1/1-(-1)=1/2灞炰簬S 3锛夎嫢S鏄崟鍏冪礌闆嗭紝鍒檃=1/1-a锛宎^2-a+1=0 鍥犱负螖=1-4=-3锛0锛屾墍浠ユ棤瀹炴暟瑙ｃ係涓嶈兘鏄崟鍏冪礌闆 ...
绛旓細A閫夐」 鍥犱负AC=14 B=30 鎵浠 BC鈮14sin30掳=7 鑰岄涓 a=7 鍙湁涓瑙 B閫夐」 浠鐐逛负鍦嗗績 r=30涓哄崐寰勭敾鍦嗭紝鍙湁涓涓氦鐐 C閫夐」 b=9 A=45掳 鎵浠 a鈮9鈭2/2=6.3 鑰岄涓璦=6 鏃犺В D閫夐」 杩欐槸涓瀹氬瓨鍦ㄧ殑锛屾牴鎹钩琛屽洓杈瑰舰娉曞垯鐢 8.3b=2鈭3asinB 姝ｅ鸡瀹氱悊...
绛旓細鍥炵瓟锛氱涓棰樻棤瑙,鍥犱负娌℃湁璇存槑杩欐槸浠涔堟暟鍒,杩欎袱涓潯浠朵笉瓒充互姹傚嚭绛旀銆 绗簩棰樺洜涓哄凡缁忕煡閬撴槸绛夊樊鏁板垪,abc,鍏朵腑a+c=2b,a+b+c=9,寰楀嚭b=3,閭ｄ箞a+c=6,a=6-c,a^2+b^2+c^2=(6-c)^2+9+c^2=35,浠庤屽緱鍑篶^2-6c+5=0,瑙ｅ緱c=1鎴5,楠岃瘉涓涓嬪氨鐭ラ亾,c=1绗﹀悎鏉′欢,鎵浠=5,c=...
