题目内容
计算:(1)14| 2 |
| 3 |
| 1 |
| 3 |
| 1 |
| 3 |
(3)(x-2)2(x+2)2•(x2+4)2. (4)(5x+3y)(3y-5x)-(4x-y)(4y+x).
分析:(1)将算式化为(15-
)(15+
),利用平方差公式计算;
(2)系数相乘,同底数幂相乘;
(3)利用a2b2=(ab)2,平方差公式进行计算;
(4)先平方差公式计算,再合并同类项.
| 1 |
| 3 |
| 1 |
| 3 |
(2)系数相乘,同底数幂相乘;
(3)利用a2b2=(ab)2,平方差公式进行计算;
(4)先平方差公式计算,再合并同类项.
解答:解:(1)14
×15
,
=(15-
)(15+
),
=152-(
)2,
=224
;
(2)-12x3y4÷(-3x2y3)•(-
xy),
=-12÷3×
x3-2+1y4-3+1,
=-
x2y2;
(3)(x-2)2(x+2)2•(x2+4)2,
=[(x-2)(x+2)]2•(x2+4)2,
=(x2-4)2•(x2+4)2,
=(x4-16)2,
=x8-32x4+256;
(4)(5x+3y)(3y-5x)-(4x-y)(4y+x),
=(9y2-25x2)-(4x2+15xy-4y2),
=-29x2-15xy+13y2.
| 2 |
| 3 |
| 1 |
| 3 |
=(15-
| 1 |
| 3 |
| 1 |
| 3 |
=152-(
| 1 |
| 3 |
=224
| 8 |
| 9 |
(2)-12x3y4÷(-3x2y3)•(-
| 1 |
| 3 |
=-12÷3×
| 1 |
| 3 |
=-
| 4 |
| 3 |
(3)(x-2)2(x+2)2•(x2+4)2,
=[(x-2)(x+2)]2•(x2+4)2,
=(x2-4)2•(x2+4)2,
=(x4-16)2,
=x8-32x4+256;
(4)(5x+3y)(3y-5x)-(4x-y)(4y+x),
=(9y2-25x2)-(4x2+15xy-4y2),
=-29x2-15xy+13y2.
点评:本题考查了单项式、多项式的乘法法则,乘法公式的灵活运用及混合运算的问题.
练习册系列答案
相关题目