圆内接正七边形A1A2A3A4A5A6A7,证明:1/A1A2=1/A1A3+1/A1A4

圆内接正七边形A1A2A3A4A5A6A7,证明:1/A1A2=1/A1A3+1/A1A4 已知A1A2A3A4A5A6A7是圆内接正七边形,求证:1/边A1A2=1/边A1A3+1/边A1A4 已知A1A2A3A4A5A6A7是圆内接正七边形,求证:1/A1A2=1/A1A3+1/A1A4 设A1A2A3A4A5A6A7是圆内接正七边形,求证：1/A1A2=1/A1A3+1/A1A4 . 已知A1A2A3A4A5A6A7是圆内接正七边形,求证;1/A1A2=1/A1A3+1/A1A4 设A1A2A3.A7是圆内接正七边形,求证:1/(A1A2)等于1/(A1A3)+1/(A1A4) 数学归纳法证明(a1+a2+.+an)^2=a1^2+a2^2+.+an^2+2(a1a2+a1a3+.+a(n-1)* 设a1,a2,a3都不为0,若1/a1a2+1/a2a3=2/a1a3,证明a1,a2,a3成等差数列 设入1入2 是矩阵A的两个不同的特征值,a1a2 分别属于特征值入1入2 的特征向量,证明：a1a2 线性无关 已知数列{1/an}为等差数列,且a1a3+a3a5+a5a1=3/5,a1a3a5=1/15,求a3 等比数列{an},满足a1+a4=18,a1a4=32,an+1>an,求{an}通项公式与Sn(a后面的都是下标）详解 （2014•烟台二模）定义a1a2a3a4=a1a4-a2a3，若f（x）=sin(π−x)3cos(π+x)1，则f（