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X-7 = x-4, then x is a number, and the absolute value of those rational numbers is smaller than itself most urgent
When X-7, X-7, X-7, X-7
It is known that the absolute value of (x + Y-Z) of rational numbers x, y and Z satisfying the formula (x-4) is 0
To the power of (13 / 10 x + 3y-3z)
(x-4)^2+(1/4)*|x+y-z|=0
The absolute value of the sum of squares is greater than or equal to 0, and the sum is equal to 0. If one is greater than 0, then the other is less than 0
So both are equal to zero
So x-4 = 0, x + Y-Z = 0
x=4,y-z=-x=
So 5x + 3y-3z = 5x-3 (Y-Z) = 5x + 3 * (- x) = 5x-3x = 2x = 2 * 4 = 8
So (5x + 3y-3z) ^ 2003 = 8 ^ 2003
That is to find the end number of 8 ^ 2003
8 ^ 1 the last mantissa is 8
8 ^ 2 ends with 4
8 ^ 3 ends with 2
The last mantissa of 8 ^ 4 is 6
8 ^ 5 the last mantissa is 8
The last mantissa of 8 ^ 6 is 4
……
So it's a cycle of four
The remainder of 2003 / 4 is 3
So the end number of 8 ^ 2003 is the same as that of 8 ^ 3
So the last mantissa of (5x + 3y-3z) 2003 is 2
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