已知丨a丨=3,b=2,且丨a-b丨=b-a,求代数式9a²-[7(a²-2/7·b)-3(1/3a²-b)-1]的值.
问题描述:
已知丨a丨=3,b=2,且丨a-b丨=b-a,求代数式9a²-[7(a²-2/7·b)-3(1/3a²-b)-1]的值.
答
∵丨a丨=3,b=2,且丨a-b丨=b-a
∴ a=-3 b=2
∴9a²-[7(a²-2/7·b)-3(1/3a²-b)-1]
=9a²-7(a²-2/7·b)+3(1/3a²-b)+1
=9a²-7a²+2b+a²-3b+1
=3a²-b+1
=3*(-3)²-2+1
=27-2+1
=26
答
因为 丨a-b丨=b-a>0,所以b>a,因为b=2,所以a=-3
将a,b的值代入代数式就可以得出答案了.