试题

题目:
计算:
(1)(2a-5)(-2a-5);
(2)(-
1
3
a+
1
2
b)(-
1
3
a-
1
2
b)

(3)(5ab-3x)(-3x-5ab);
(4)(
1
2
x-2)(
1
2
x+2)-
1
4
x(x+8)

(5)(x-y)(
1
9
x-y)-(
1
3
x-y)(
1
3
x+y)

答案
解:(1)(2a-5)(-2a-5)=25-4a2

(2)(-
1
3
a
+
1
2
b
)(-
1
3
a
-
1
2
b
)=
1
9
a2-
1
4
b2


(3)(5ab-3x)(-3x-5ab)=9x2-25a2b2

(4)(
1
2
x-2
)(
1
2
x+2
)-
1
4
x
(x+8),
=
1
4
 x2
-4-
1
4
 x2
-2x,
=-2x-4;

(5)(x-y)(
1
9
x-y
)-(
1
3
x-y
)(
1
3
x+y
),
=(x-y)(
1
9
x-y
)-(
1
9
x2-y2
),
=2y2-
10
9
xy

解:(1)(2a-5)(-2a-5)=25-4a2

(2)(-
1
3
a
+
1
2
b
)(-
1
3
a
-
1
2
b
)=
1
9
a2-
1
4
b2


(3)(5ab-3x)(-3x-5ab)=9x2-25a2b2

(4)(
1
2
x-2
)(
1
2
x+2
)-
1
4
x
(x+8),
=
1
4
 x2
-4-
1
4
 x2
-2x,
=-2x-4;

(5)(x-y)(
1
9
x-y
)-(
1
3
x-y
)(
1
3
x+y
),
=(x-y)(
1
9
x-y
)-(
1
9
x2-y2
),
=2y2-
10
9
xy
考点梳理
平方差公式.
根据平方差公式(a+b)(a-b)=a2-b2,即可解答本题.
本题考查了平方差公式,属于基础题,关键是掌握平方差公式(a+b)(a-b)=a2-b2
计算题.
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