试题

题目:
计算或解方程或化简求值:
(如)(x-8y)(x-y)
(2)(25x2+如5xsy-20x7)÷(-5x2
(s)|-
2
|+
(-
2
)
2
+
s-8

(7)(2x+sy)2-(2x+y)(2x-y)
(5)解方程:8x-(x+5)(x-5)=-2-(x+如)(x+s)
(6)先化简,再求值:[(x2+y2)-(x-y)2+2y(x-y)]÷7y,其中x=s,y=-7.
答案
解:(七)原式=v2-vy-8vy+8y2
=v2-9vy+8y2

(2)解:原式=25v2÷(-5v2)+七5vy÷(-5v2)-20v4÷(-5v2),
=-5-右vy+4v2

(右)解:原式=
2
+
2
-2,
=-七;

(4)解:原式=4v2+七2vy+9y2-4v2+y2
=七2vy+七0y2

(5)解:去括号得:8v-v2+25=-2-v2-v-右v-右,
移项得:8v-v2+v2+v+右v=-2-右-25,
合并同类项得:七2v=-右0,
系数化成七得:v=-
5
2


(6)解:当v=右,y=-4时,
[(v2+y2)-(v-y)2+2y(v-y)]÷4y,
=[v2+y2-(v2-2vy+y2)+2vy-2y2]÷4y,
=[v2+y2-v2+2vy-y2+2vy-2y2]÷4y,
=[-2y2+4vy]÷4y,
=-
2
y+v,
=-
2
×(-4)+右,
=5.
解:(七)原式=v2-vy-8vy+8y2
=v2-9vy+8y2

(2)解:原式=25v2÷(-5v2)+七5vy÷(-5v2)-20v4÷(-5v2),
=-5-右vy+4v2

(右)解:原式=
2
+
2
-2,
=-七;

(4)解:原式=4v2+七2vy+9y2-4v2+y2
=七2vy+七0y2

(5)解:去括号得:8v-v2+25=-2-v2-v-右v-右,
移项得:8v-v2+v2+v+右v=-2-右-25,
合并同类项得:七2v=-右0,
系数化成七得:v=-
5
2


(6)解:当v=右,y=-4时,
[(v2+y2)-(v-y)2+2y(v-y)]÷4y,
=[v2+y2-(v2-2vy+y2)+2vy-2y2]÷4y,
=[v2+y2-v2+2vy-y2+2vy-2y2]÷4y,
=[-2y2+4vy]÷4y,
=-
2
y+v,
=-
2
×(-4)+右,
=5.
考点梳理
整式的混合运算—化简求值;实数的运算;整式的混合运算.
(1)根据多项式乘多项式法则展开得出x2-xy-8xy+8y2,再合并同类项即可;
(2)根据多项式除以单项式法则先展开得出25x2÷(-5x2)+15x3y÷(-5x2)-20x4÷(-5x2),再根据单项式除以单项式法则进行计算即可;
(3)求出每一部分的值,再代入求出即可;
(4)根据平方差公式和完全平方公式先展开得出4x2+12xy+9y2-4x2+y2,再合并同类项即可;
(5)去括号得出8x-x2+25=-2-x2-x-3x-3,移项、合并同类项得出12x=-30,系数化成1即可;
(6)先算小括号里面的,在去小括号后合并同类项得出[-2y2+4xy]÷4y,推出-
1
2
y+x,代入求出即可.
本题主要考查了整式的运算,平方差公式和完全平方公式,解一元一次方程等知识点的应用,考查了学生综合运用性质进行计算的能力,题目都比较典型,但是一些比较容易出错的题目.
计算题.
找相似题