七年级下数学化简求值练习题

1、已知A,B是方程x^2+2x-5=0的两个实数根,

求(A^2+2AB+2A)(B^2+2AB+2B)的值.

由A,B是方程x^2+2x-5=0的两个实数根得：

AB=-5，A+B=-2

A^2+2AB+2A)(B^2+2AB+2B)

=AB（A+2B+2）（B+2A+2）

=-5（-2+B+2）（-2+A+2）

=-5AB

=25

2、1/2(x+y+z)方+1/2(x-y-z)(x-y+z)-z(x+y),其中x-y=6,xy=21.要详细步骤

化简得：

1/2(x+y+z)方+1/2(x-y-z)(x-y+z)-z(x+y)=

1/2[(x+y)方+2z(x+y)+z方]+1/2[(x-y)方-z方]-z(x+y)=

1/2(x+y)方+1/2(x-y)方=x方＋y方

由x-y=6,xy=21得，x方＋y方＝（x-y）方＋2xy＝78

3、a^2-ab+2b^2=3求2ab-2a^2-4b^2-7的值

2ab-2a^2-4b^2-7

=2(ab-a^2-2b^2)-7

=-2(a^2-ab+2b^2)-7

=(-2)*3-7

=-6-7=-13

4、若A=2x^2+3xy-2x-3,B=-x^2+xy+2,且3A+6B的值与x无关，求y的值

解：

3A+6B=6x^2+9xy-6x-9-6x^2+6xy+12

=15xy-6x+3

=x(15y-6)+3

5、9x+6x^2-3(x-2/3x^2).其中x=-2

9x+6x�0�5-3(x-2/3x�0�5)

=9x+6x�0�5-3x+2x�0�5

=8x�0�5+6x

=8×(-2)�0�5+6×(-2)

=32-12

=20

6、1/4(-4x^2+2x-8)-(1/2x-1),其中x=1/2

1/4(-4x�0�5+2x-8)-(1/2x-1)

=-x�0�5+1/2x-2-1/2x+1

=-x�0�5-1

=-(1/2)�0�5-1

=-1/4-1

=-5/4

7、3x'y-[2x'y-(2xyz-x'z)-4x'z]-xyz,其中x=-2,y=-3,z=1,

:3x'y-[2x'y-(2xyz-x'z)-4x'z]-xyz

=3x'y-2x'y+2xyz-x'z+4x'z-xyz

=x'y-xyz+3x'z

=4*(-3)-2*3*1+3*4*1

=-12-6+12

=-6

8、(5a^2-3b^2)+(a^2+b^2)-(5a^2+3b^2),其中a=-1,b=1

=5a^2-3b^2+a^2+b^2-5a^2-3b^2

=a^2-5b^2

=(-1)^2-5*1^2

=1-5

=-4

9、2(a^2b+ab^2)-2(a^2b-1)-2ab^2-2其中a=-2,b=2

=2a^2b+2ab^2-2a^2b+2-2ab^2-2

=0

10、(X-2分之1Y-1)(X-2分之1Y+1)-(X-2分之1Y-1)的平方

其中X=1.7,Y=3.9(先化简再求值)

[(X-2分之1Y)-1][(X+2分之1Y)+1]-(X-2分之1Y-1)平方

=(X+2分之1Y)平方-1-(X-2分之1Y)平方+2(X-2分之1Y)-1

=(X+2分之1Y)平方-(X-2分之1Y)平方+2(X-2分之1Y)-2

=2XY+2X-Y-2

=3.9*2.4+1.4

=10.76