题目内容
计算下列各题:
(1)
;
(2)[(2x+1)(4x+2)-2]÷(-8x);
(3)(3x+1)2﹣(3x-1)(3x+1)
(4)( a-b-c )2
(5)已知:x+y=-1,xy=-6,求:x2+y2及x-y的值.
(1)
(2)[(2x+1)(4x+2)-2]÷(-8x);
(3)(3x+1)2﹣(3x-1)(3x+1)
(4)( a-b-c )2
(5)已知:x+y=-1,xy=-6,求:x2+y2及x-y的值.
解:
(1)原式=﹣1﹣
×4×1=﹣1.5;
(2)原式=(8x2﹣8x+2﹣2)÷(﹣8x)=﹣x﹣1;
(3)原式=[(x+2y)(x﹣2y)]2﹣(x﹣2y)(x+2y)(x2+4y2)
=(x2﹣4y2)2﹣(x2﹣4y2)(x2+4y2)
=x4﹣8x2y2+16y4﹣x4+16y4
=32y4﹣8x2y2
(4)原式=(a﹣b)2﹣2ac+4bc+c2
=a2﹣4ab+4b2﹣2ac+4bc+c2
(5)原式=(x+y)2﹣2xy=13;
原式=±
=±5.
(1)原式=﹣1﹣
(2)原式=(8x2﹣8x+2﹣2)÷(﹣8x)=﹣x﹣1;
(3)原式=[(x+2y)(x﹣2y)]2﹣(x﹣2y)(x+2y)(x2+4y2)
=(x2﹣4y2)2﹣(x2﹣4y2)(x2+4y2)
=x4﹣8x2y2+16y4﹣x4+16y4
=32y4﹣8x2y2
(4)原式=(a﹣b)2﹣2ac+4bc+c2
=a2﹣4ab+4b2﹣2ac+4bc+c2
(5)原式=(x+y)2﹣2xy=13;
原式=±
练习册系列答案
相关题目