几道简单的整式计算,初一的

来源：百度知道 编辑：UC知道 时间：2024/06/05 15:59:10
尽量写清楚一点

计算:
(1-x)(0.6-x)

(2x+y)(x-y)

(2n+5)(n-3)

(ax+b)(cx+d)

化简(x-y)(x-2y)- (1/2) (2x-3y)(x+2y), 并求出当x=2, y=5时的值

解：原式=0.6-x-0.6x+x2
=x2-1.6x+0.6
原式=2x2-2xy+xy-y2
=2x2-xy-y2
原式=2n2-6n+5n-15
=2n2-n-15
原式=acx2+adx+bcx+bd
=acx2+(ad+bc)x+bd
原式=(x2-2xy-xy+2y2)-(1/2)(2x2+4xy-3xy-6y2)
=(x2-3xy+2y2)-(1/2)(2x2+xy-6y2)
=x2-3xy+2y2-x2-(1/2)xy+3y2
=-(7/2)xy+5y2
将x=2, y=5带入得：-(7/2)*2*5+5*5*5
=90

(1)0.6-1.6x+x^2
(2)2x^2-xy-y^2
(3)2n^2-n-15
(4)acx^2+(ad+bc)x+bd
(5)原式=(x^2-3xy+2y^2)-(1/2) (2x^2+xy-6y^2)
=-(7/2)xy+5y^2
当x=2, y=5时,原式=-(7/2)*2*5+5*5^2
=90

(1-x)(0.6-x)
=0.6-X-0.6X+X^
=X^-(1+0.6)X+0.6
=X^-1.6X+0.6

(2x+y)(x-y)
=2x^-2xy+xy-y^
=2x^-(2-1)xy-y^
=2x^-xy-y^

(2n+5)(n-3)
=2n^-6n+5n-15
=2n^-(6-5)n-15
=2n^-n-15

(ax+b)(cx+d)
=ax^+adx+bcx+bd
=ax^+(ad+bc)x+bd

(x-y)(x-2y)- (1/2) (2x-3y)(x