x>y>1 (y!)^(x-1) 和(x!)^(y-1)的大小
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x>y>1 (y!)^(x-1) 和(x!)^(y-1)的大小
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![x>y>1 (y!)^(x-1) 和(x!)^(y-1)的大小](/uploads/image/z/18794440-64-0.jpg?t=x%3Ey%3E1+%28y%21%29%5E%28x-1%29+%E5%92%8C%28x%21%29%5E%28y-1%29%E7%9A%84%E5%A4%A7%E5%B0%8F)
(y!)^(x-1) 和(x!)^(y-1)
题目出现阶乘“!”,所以x、y为自然数.
又x>y>1,所以y≥2,x≥3
不妨设x=y+k,k为自然数
即:
[(y!)^(x-1) ]/[(x!)^(y-1)]=[(y!)^(y+k-1)]/{[(y+k)!]^(y-1)}
=[(y!)^(y-1+k)]/{[(y!(y+1)(y+2)……(y+k)]^(y-1)}
=[(y!)^(y-1+k)]/{[(y!)^(y-1)][(y+1)(y+2)……(y+k)]^(y-1)}
=[(y!)^(y-1)][(y!)^k]/{[(y!)^(y-1)][(y+1)(y+2)……(y+k)]^(y-1)}
=[(y!)^k]/{[(y+1)(y+2)……(y+k)]^(y-1)}
<[(y!)^k]/{[(y+1)^k]^(y-1)}
=[(y!)^k]/{[(y+1)^k]^(y-1)}
<[(y!)^k]/[(y^k)^(y-1)]
=[(y!)^k]/[y^(y-1)]^k
=[(y!/y^(y-1)]^k
={(y/y)[(y-1)/y][(y-2)/y]……[3/y][2/y]*1}^k
<1^k=1
所以[(y!)^(x-1) ]/[(x!)^(y-1)]<1
则(y!)^(x-1)<(x!)^(y-1).
题目出现阶乘“!”,所以x、y为自然数.
又x>y>1,所以y≥2,x≥3
不妨设x=y+k,k为自然数
即:
[(y!)^(x-1) ]/[(x!)^(y-1)]=[(y!)^(y+k-1)]/{[(y+k)!]^(y-1)}
=[(y!)^(y-1+k)]/{[(y!(y+1)(y+2)……(y+k)]^(y-1)}
=[(y!)^(y-1+k)]/{[(y!)^(y-1)][(y+1)(y+2)……(y+k)]^(y-1)}
=[(y!)^(y-1)][(y!)^k]/{[(y!)^(y-1)][(y+1)(y+2)……(y+k)]^(y-1)}
=[(y!)^k]/{[(y+1)(y+2)……(y+k)]^(y-1)}
<[(y!)^k]/{[(y+1)^k]^(y-1)}
=[(y!)^k]/{[(y+1)^k]^(y-1)}
<[(y!)^k]/[(y^k)^(y-1)]
=[(y!)^k]/[y^(y-1)]^k
=[(y!/y^(y-1)]^k
={(y/y)[(y-1)/y][(y-2)/y]……[3/y][2/y]*1}^k
<1^k=1
所以[(y!)^(x-1) ]/[(x!)^(y-1)]<1
则(y!)^(x-1)<(x!)^(y-1).
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