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小专题7整式的运算和化简求值
类型一整式的化简
1.计算:
(1)3m2·m4-(2m2)3+4m8÷m2:
(2)(2x2y)2·(-5y2)÷(10xy3)+5xy
2.计算:
(1)[(-a-b)2+(a-b)(-a-b)]÷(-2b):
(2)(易错题)(x+y+z)(x+y-z)-(x+y+z)2.
类型二利用直接代入法化简求值
3.先化简,再求值:
()(3a-1)2-6a(a-1),其中a=-写
(2)(2x+y)(x-4y)+(2x-4xy-3x3y2)÷(-x)3,其中x=-2,y=-3;
83
(3)(2a-b)2-(a-2b)(a+2b)+(6a2b+8ab2)÷2b,其中a=2,b=-1.
类型三利用条件间接代入法化简求值
4.先化简,再求值:(m-n)(m+n)-(m-n)2+2n2,其中12m-11+(n+1)2=0.
5.先化简,再求值:[(x+2y)(x-2y)-(2x-y)2-(x2-5y2)]÷(-2x),其中x,y满
足2m÷2y=8.
类型四利用整体代入法化简求值
6.先化简,再求值:(2+o)2-a)+a(a-56)+3a6÷(-bP,其中山=-2
7.已知5x2-x-1=0,求代数式(3x+2)(3x-2)+x(x-2)的值,
84=x-81y2
8.解:(1)(1-x)(1+x)=1-x2:
(2)原式=[(3x-2y)·(3x+2yr)]2
(1-x)(1+x+x2)=1-x3:
=(9x2-4y2)2
(1-x)(1+x+x2+x3)=1-x:
=81x-72x2y2+16y
...
(3)原式=[x+(y-5)][x-(y-5)]
.(1-x)(1+x+x2+x3+…+x-1)=
=x2-(y-5)2
1-x"
=x2-y2+10y-25.
(2)①(1-2)(1+2+22+23+2+23+26)
(4)原式=[(y2-4)+y][(2-4)-y]
=1-27=1-128=-127.
=(y2-4)2-y2
②(x-1)(x204+x223+x22+…+x2+x+
=y-8y2+16-y2
1)=-(1-x)(1+x+x2+…+x204)=
=y-9y2+16.
-(1-x2025)=x205-1.
3.C【解析】x+y=10,y=15,.x2-xy+y
(3)1.理由如下:20+2”+2%+…+22+2+
=(x+y)2-3y=102-3×15=55.故选C.
1=-(1-2)×(1+2+22+…+20)=
4.51【解析】小a+b=3,ab=2,a2+2=
-(1-2)=2-1.2的个位数是2,2
(a+b)2-2ab=32-2×2=5..(a-b)2=
的个位数是4,2的个位数是8,2的个位数
a2-2ab+b=5-2×2=1.
是6,2的个位数是2,…∴.2"(n≥1,且n
5解:-°=2-2…+
为整数)的个位数以2,4,8,6不断循环出现
…101÷4=25…1,,20的个位数是2
=*2-4(+
2-1的个位数是1.
小专题7整式的运算和化简求值
4=32-4=5
1.解:(1)原式=3m°-8m+4m°=-m°.
(2:(e-=-2+
(2)原式=4xy2·(-52)÷(10xy3)+5
=(-20x3y)÷(10xy3)+5.y
+=(+2=5+2=7
=-2xy +5xy
=3xy.
(+=+2+
2.解:(1)原式=(a2+2ab+b2+b2-a2)÷
(-2b)》
+(+-2=49-2=4
=(2ab+2b)÷(-2b)
6.解:(1)(a+b+c)2=a2+b2+c2+2ab+
=-a-b.
2ac +2bc.
(2)原式=(x+y)2-2-[(x+y)+]
(2).:a+b+c=11,ab+bc+ac=38,
=(x+y)2-2-[(x+y)2+2z(x+y)+2]
(a+b+c)2=a2+b2+c2+2ab+2ac+2bc,
=(x+y)2-2-(x+y)2-2:(x+y)-
a2+6+c2=121-2×38=45.
=-222-2x2-2ya
3.解:(1)原式=9a2-6a+1-6a2+6a
7.解:,A=(4n+1)(4n-1)-(n+1)(n-1)
=16m2-1-(n2-1)=16m2-1-m2+1=
=3a2+1.
15n2(n是正整数),∴15n2能被15整除.
当a=号时,原式=3×(-)+1
∴.整式A=(4n+1)(4n-1)-(n+1)(n-
1)能被15整除.
g+1=3+1=
=3×
147
(2)原式=2x2-8xy+y-4y2+(2x5-4xy6.解:(1)原式=3bc(5a-c).
-3xy2)÷(-x3)=2x2-7y-4y2-2x2+
(2)原式=2x(x-6y2+4y3)
4xy+3y2=-3.y-y