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A, b are mutually opposite numbers, c, d are mutually reciprocal, m is a negative number with an absolute value of 3, find the square of m+(cd+a+b) times the power of m+(cd) of 2008
A, b are opposite numbers, then a+b=0
C, d are reciprocal, then cd=1
M is a negative number with an absolute value of 3, then m =-3,
Therefore:
M2+(cd+a+b) m+(cd)^2008
=(-3)²+(1+0)*(-3)+(1)^2008
=9-3+1
=7
A, b are opposite numbers, then a+b=0
C, d are reciprocal, then cd=1
M is a negative number with an absolute value of 3, then m =-3,
So:
M2+(cd+a+b) m+(cd)^2008
=(-3)²+(1+0)*(-3)+(1)^2008
=9-3+1
=7
Given that a and b are mutually opposite numbers, c and d are mutually reciprocal numbers, m is a negative number whose absolute value is equal to 3, and the value of m2+(cd+a+b) multiplied by m+(-cd) to the power of 2008 is obtained. Urgent Given that a and b are mutually opposite numbers, c and d are mutually reciprocal numbers, m is a negative number whose absolute value is equal to 3, and the value of m squared +(cd + a + b) multiplied by m +(-cd) to the obtained. Urgent
[Analysis]: Calculate according to the sum of two numbers opposite to each other is 0; the product of two numbers reciprocal to each other is 1.
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[Problem Solving]:
A, b are mutually opposite numbers,
A+b=0,
C, d are reciprocal,
Cd=1,
M is a negative number with an absolute value of 3,
M=-3,
M2+(cd+a+b)×m+(-cd)^2008=9+(1+0)×(-3)+1=7.
Therefore, the answer to this question is:7.
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[Summary]: Note the overall substitution in the topic, and relevant knowledge points: the sum of two numbers opposite to each other is 0; the product of two numbers reciprocal to each other is 1; the odd power of -1 is -1, and even power is 1.
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