问题 解答题
计算
(1)(
1
100
)0+(-3)2+3-1-(-3)2

(2)已知x+
1
x
=10
,求x2+
1
x2
的值
(3)(x+6)2-(x-3)(x+3)2
(4)[(3xy+z)2-z2]÷xy.
答案

(1)原式=1+9-

1
3
-9

=

2
3

(2)∵x+

1
x
=10,

∴(x+

1
x
2=x2+2+
1
x2
=100,

则x2+

1
x2
=98;

(3)原式=x2+12x+36-(x-3)(x2+6x+9)

=x2+12x+36-(x3-27)

=x2+12x+36-x3+27;

(4)原式=(9x2y2+6xyz+z2-z2)÷xy

=9xy+6z.

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