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How many sets of integers x yz satisfying the 1999 power of the absolute value of x+y plus the 1999 power of the absolute value of x+z plus the 2000 power of the absolute value of y+z =2? How many sets of integers x yz satisfying the power of 1999 of the absolute value of x + y plus the power of 1999 of the absolute value of x + z plus the power of 2000 of the absolute value of y + z =2?
This is actually a permutation combination problem. The absolute value of the array can only be equal to 0,±1. There are and only two groups can be ±1. The other item is 0.
{0,±1,±1}{±1,0,±1}{±1,±1,0}, So there are 12 groups of integer group xyz values
This is actually a permutation and combination problem, the absolute value of the array can only be equal to 0,±1, there and only can be two groups of ±1, the other is 0,
{0,±1,±1}{±1,0,±1}{±1,±1,0}, So there are 12 groups of integer group xyz values
Given the square of x+x=1, find the value of the third power of 2004X+ the square of 2003x—2005x-2007
X2+x=1
The original formula =2004x3+2004x2-x2-2005x-2007
=2004X (x2+x)-x2-2005x-2007
=2004X*1-x2-2005x-2007
=-X2-x-2007
=-(X2+x)+2007
=-1+2007
=2006
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