。帮忙解答下高一数学题，，，谢了！！！急急急！！！答案写清楚哦
\u5f53\u5802\u68c0\u6d4b \uff0c\uff0c\u5e2e\u6211\u89e3\u7b54\u4e0b\u9ad8\u4e00\u6570\u5b66\u9898\uff0c\uff0c\uff0c\u8c22\u4e86\uff0c\uff0c\u6025\u6025\u6025\u6025\uff01\uff01\uff01\u7b54\u6848\u5199\u6e05\u695a\u54e6\uff01\uff011A,2A,3A,4d=4,5a=33,b=18,c=3
B\u5c5eA\uff0c\u90a3\u4e48a²=4\u6216a
a²=4\u65f6\uff0ca=2\u6216\u3010-2\u3011
a²=a\u65f6\uff0ca=0\u62161 \u30101\u4e0d\u5408\u9898\u610f\uff0c\u820d\u3011
由cosa=(b²+c²-a²)/2bc
有
(b²-2)/2根号2b=根号2/4
b²-b-2=0
(b-2)(b+1)=0
b=2
三角形为等腰三角形，a和b等腰
sinA=根号(1-cos²A)=(根号14)/4
sin2A=2sinAcosA=根号7/4
cos2A=3/4
2.
cos(2A+π/3)=cos2Acosπ/3-sin2Asinπ/3=3/4*1/2-根号7/4*根号3/2=(3-根号21)/8
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1.
由cosa=(b²+c²-a²)/2bc
有
(b²-2)/2根号2b=根号2/4
b²-b-2=0
(b-2)(b+1)=0
b=2
三角形为等腰三角形，a和b等腰
sinA=根号(1-cos²A)=(根号14)/4
sin2A=2sinAcosA=根号7/4
cos2A=3/4
2.
cos(2A+π/3)=cos2Acosπ/3-sin2Asinπ/3=3/4*1/2-根号7/4*根号3/2=(3-根号21)/8
1)cosA=b^2+c^2-a^2/(2bc),代入a、c求b
sin^2 A = 1 - cos^2 A,取正
2)利于两角余弦展开公式，求解
coscos-sinsin
绛旓細鍥
绛旓細濡傚浘
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绛旓細鍒橲n=na1+n(n-1)d/2 =12n+n(n-1)x2/2 =n²+11n =242 鍗硁²+11n-242=0 (n+22)(n-11)=0 瑙ｅ緱锛歯=11鎴-22(鑸嶅幓)鍒檔=11 銆鏁板鐨勫揩涔愩戝洟闃熶负鎮瑙ｇ瓟锛佺鎮ㄥ涔犺繘姝 涓嶆槑鐧藉彲浠ヨ拷闂紒婊℃剰璇风偣鍑讳笅闈㈢殑銆愰変负婊℃剰鍥炵瓟銆戞寜閽紝O(鈭鈭)O璋㈣阿 ...
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绛旓細f(x)=2sin(pi/2-x-pi/4)cos(x-pi/4)+3^0.5sin2x=2sin(pi/4-x)cos(x-pi/4)+3^0.5sin2x =-sin[2(x-pi/4)]+3^0.5sin2x=-sin(2x-pi/2)+3^0.5sin2x=cos2x+3^0.5sin2x=[1^2+(3^0.5)^2]sin(2x+pi/6)=2sin(2x+pi/6)1.-1<=sin(2x+pi/6)<=1,鎵浠...
绛旓細鎴戣寰楀氨绗4涓瘮杈冮夯鐑︼紝鍏堟眰鍑洪氶」鍏紡an=鏍瑰彿銆愶紙100^n-1锛/9 -- (10^n-1)/9*2銆=(10^n-1)/3锛孲n=鏍瑰彿(11鈥2)+ 鏍瑰彿(1111鈥22)鈥+鏍瑰彿(11鈥11鈥22鈥2)=锛10^1-1)/3+锛10^2-1)/3+鈥︹︹︼紙10^n-1)/3锛=锛10*10^n-10锛/27 - 1/3 *n.鈶犵敤閿欎綅鐩稿噺锛歛S...
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=...
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...
