问题 解答题
(1)(-1)-2+(π-23)0-(-
1
2
)-1

(2)已知:a+
1
a
=3,求a2+
1
a2
的值.
(3)(9x3y2-6x2y+3xy2)÷(-3xy);
(4)先化简再求值:(m+1)2-5(m+1)(m-1)+3(m-1)2,其中m=-
1
2
答案

(1)(-1)-2+(π-23)0-(-

1
2
)-1

=1+1+2

=4;

(2)a2+

1
a2
=(a+
1
a
2-2=32-2=7;

(3)(9x3y2-6x2y+3xy2)÷(-3xy)

=-3x2y+2x-y;

(4)(m+1)2-5(m+1)(m-1)+3(m-1)2

=m2+2m+1-5m2+5+3m2-6m+3

=-m2-4m+9,

m=-

1
2
时,原式=-
1
4
+2+9=10
3
4

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