设函数f(x)=2x-cosx,an是公差为八分之pai的等差数列,f(a1)+f(a2)+…+f(a5)等于5pai,则f(a3)的平方减a1*a3等于几

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设函数f(x)=2x-cosx,an是公差为八分之pai的等差数列,f(a1)+f(a2)+…+f(a5)等于5pai,则f(a3)的平方减a1*a3等于几

设函数f(x)=2x-cosx,an是公差为八分之pai的等差数列,f(a1)+f(a2)+…+f(a5)等于5pai,则f(a3)的平方减a1*a3等于几

设函数f(x)=2x-cosx,an是公差为八分之pai的等差数列,f(a1)+f(a2)+…+f(a5)等于5pai,则f(a3)的平方减a1*a3等于几

设函数f(x)=2x-cosx,an是公差为八分之pai的等差数列,f(a1)+f(a2)+…+f(a5)等于5pai,则f(a3)的平方减a1*a3等于几

设函数f(x)=sinx+cosx,f'(x)是f(x)的导函数,若f(x)=2f'(x)求[(sinx)^2-sin2x]/(cosx)^2 设函数f(x)=2x-cosx,【an】是公差为π/8的等差数列,f(a1)+f(a2)+..+f(a5)=5π则【f(a3)】²-a1a5=求详解 10、设函数f(x)=2x-cosx,{an}是公差为π/8的等差数列,f(a1)+f(a2)+…+f(a5)=5π10、设函数f(x)=2x-cosx,{an}是公差为π/8的等差数列,f(a1)+f(a2)+…+f(a5)=5π,则[f(a3)]2-a1a5=______.A 0 B 1/16*π2 设函数f(x){xe^(x^2),x>=0 {1/cosx ,-π 设函数fx=2x-cosx,{an}是公差为π/8的等差数列 ,f(a1)+f(a2)设函数f(x)=2x-cosx,{an}是公差为π/8的等差数列,f(a1)+f(a2)+…f(a5)=5π,则[f(a)]^2-a1*a5= 设函数f(x)=2x-cosx,an是公差为π的等差数列,f(a1)+f(a2)+f(a3)=3π,则f(a1)+f(a2)+f(a3)+……+f(a10)= 函数f(x)=x+cosx的一个原函数是?设函数f(x)=xarcsinx,则f’(x)=? 设函数f(x)=2x-cosx,{An}是公差为(π/8)的等差数列,f(a1)+f(a2)+…f(a5)=5π,则[f(a)]^2-a1*a3= 设函数f(x)=2x-cosx,{an}是公差为π/8的等差数列,f(a1)+f(a2)+…f(a5)=5π,则[f(a)]^2-a2*a3= 设函数f(x)=2x-cosx,{An}是公差为TT/8的等差数列,f(a1)+f(a2)+…f(a5)=5TT,则 f[(a3)]^2-a1a3= 设函数f(x)=2x-cosx,{An}是公差为π的等差数列,f(a1)+f(a2)+…f(a5)=5π,则[f(a)]^2-a1*a3= 设函数f(x)=sinx-cosx+x+1,0 设函数f(x)=sinx-cosx+x+1,0 设f(cosx-1)=cosx^2,求f(x) 设分段函数f(x)=sinx(sinx>=cosx),cosx(sinx 已知函数f(x)=x²/x²+1,设f(n)=an(n∈N+)(1)求证:an>1(2){an}是递增数列还是递减数列已知函数f(x)=x²/x²+1,设f(n)=an(n∈N+)(1)求证:an>1(2){an}是递增数列还是递减数列 设an是函数f(x)=x^3+n^2*x-1的零点,证明;a1+a2+..an 设函数f(x)=(2sinxcosx+5/2)/(sinx+cosx),[0 ,π/2] 求f(x)的最小值