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规定A x m =x（x1）（xm1），其中x∈R，m为正整数，且A x 0 =1，这是排列数A n m （n，m是正整数，且m≤n）的一种推广． （1）求A15 3 的值； （2）排列数的两个性质：①A n m =nA n1 m1 ，②A n m mA n m1 =A n1 m ．（其中m，n是正整数）是否都能推广到A x m （x∈R，m是正整数）的情形？ 若能推广，写出解；∵A x m =x（x1）（xm1），其中x∈R，m为正整数．∴ 3 =（9）×（10）×（11）=990 故答案为990 本回答由提问者推荐Y dx M x x dx M x x x dx x x x M M 1 51 2 151 512 256 2 1515 x x M M from CE m23 at Mapúa Institute of Technology

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ƒAƒxƒm‹´–‚–@¤"XŠX op-Solve your math problems using our free math solver with stepbystep solutions Our math solver supports basic math, prealgebra, algebra, trigonometry, calculus and moreSolve your math problems using our free math solver with stepbystep solutions Our math solver supports basic math, prealgebra, algebra, trigonometry, calculus and more

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X M x { X ӂ̊ό X b g L O B x M x { X ӂɂ́uJR x w m R ~ _ F38 i5 _ _ j B n v u g ^ r e B T r X @ x X m R ~ _ F50 i5 _ _ j n v Ȃǂ ܂ B x M x { X ӂ̃z e ^ ό X b g ^ C x g ^ n O B规定A x m =x（x1）（x2）••（xm1），其中x∈R，m∈N*． 函数f（x）=aA x1 3 3bA x 2 1（ab≠0）在x=1处取得极值，在x=2处的切线的平行向量为 OP =(b5，5a)规定A x m =x（x1）（xm1），其中x∈R，m为正整数，且A x =1，这是排列数A n m （n，m是正整数，且m≤n）的一种推广． （1）求A15 3 的值； （2）排列数的两个性质：①A n m =nA n1 m1 ，②A n m mA n m1 =A n1 m ．（其中m，n是正整数）是否都能推广到A x m （x∈R，m是正整数）的情形？ 若能推广，写出推广

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