高中数学
若(x^2+1)(x-2)^9=a0+a1(x-1)+a2(x-1)^2+...+a11(x-1)^11,则(a1+3a3+..+11a11)^2-(2a2+4a4+...+10a10)^2= 请给出详细过程
(a1+3a3+..+11a11)^2-(2a2+4a4+...+10a10)^2 = (a1-2a2+3a3-...+11a11)(a1+2a2+3a3+...+11a11) ...(1) 两边微分:(x^2+1)(x-2)^9=a0+a1(x-1)+a2(x-1)^2+...+a11(x-1)^11 得:(11x^2-4x+9)(x-2)^8=a1+2a2(x-1)+...+11a11(x-1)^10 ...(2) x=2时,(2)为:0 =a1+2a2+3a3+...+11a11 ...(3) (1)(3) ==> (a1+3a3+..+11a11)^2-(2a2+4a4+...+10a10)^2 =0
答:已知f(x)=e^(-x)-ax^2+x-1(a≥0) 〔1〕当a=0时,求f(x)的单调区间及极值。〔2〕若x≤0时,恒有f(x)≥0成立,求a的取值范围 (...详情>>
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