Question

解四道初一的数学计算题

1） [(X+3)^2+(x+3)(x-3)]/2x
2) (x+1)^2+2(1-x)-x^2
3) (a-1)^2(a-1)^2(a^2+1)^2
4) (x-2y-1)(2y-x-1)
需要计算过程

Answer 1

1） [(X+3)^2+(x+3)(x-3)]/2x=（x^2+6x+9+x^2-9）/2x=(2x^2+3x)/2x=x+3
2) (x+1)^2+2(1-x)-x^2=x^2+2x+1+2-2x-x^2=3
3) (a-1)^2(a-1)^2(a^2+1)^2 你第二题是不是打错了，应是
(a-1)^2(a+1)^2(a^2+1)^2=(a^2-1)^2(a^2+1)^2=(a^4-1)^2=a^8-2a^4+1
4) (x-2y-1)(2y-x-1)=-[(x-2y)-1][(x-2y)+1]=1-(x-2y)^2=1-x^2+4xy-4y^2
这种题要背会几个基本公式就行了，要多联系。这四道题就用了两个公式
(a+1)(a-1)=a^2-1;(a-1)^2=a^2-2a+1.好好学习哦，我上学的时候怎么没有这种东西呢!hh

Answer 2

1）原式=（X^2+6X+9+X^2-9)/2X=(2X^2+6X)/2X=X+3
2)原式=X^2+2X+1+2-2X-X^2=3
3）原式=[(A-1)(A+1)]^2(A^2+1)^2=(A^2-1)^2(A^2+1)^2=[(A^2-1)(A^2+1)]^2=(A^4-1)^2=A^8-2A^4+1
4)原式=[(X-2Y)-1][-(X-2Y)-1]=1-(X-2Y)^2=1-X^2+4XY-4Y^2

Answer 3

1） [(X+3)^2+(x+3)(x-3)]/2x = [x²+6x+9 + x²-9]/2x =[2x²+6x]/2x=x+3
2) (x+1)^2+2(1-x)-x^2 =x²+2x+1 + 2-2x -x² = 3
3) (a-1)^2(a-1)^2(a^2+1)^2 你是不是打错了 应该是 (a+1)^2(a-1)^2(a^2+1)^2
(a+1)^2(a-1)^2(a^2+1)^2 =[(a+1)(a-1)]²(a^2+1)^2=[a²-1]²(a^2+1)^2=[(a²-1)(a²+1)]²
=[a^4-1]²=a^8-2a^4+1
4) (x-2y-1)(2y-x-1)=[(x-2y)-1][-(x-2y)-1]=[-1+(x-2y)][-1-(x-2y)]=(-1)²-(x-2y)²
=1-(x²+4xy+4y²)=1-x²-4xy-4y²)

Answer 4

1） [(X+3)^2+(x+3)(x-3)]/2x
=（x+3)(x+3+x-3)/2x
=(x+3)*2x/2x
=x+3
2) (x+1)^2+2(1-x)-x^2
=x^2+2x+1+2-2x-x^2
=3
3) (a+1)^2(a-1)^2(a^2+1)^2
=[(a-1)(a-1)(a^2+1)]^2
=[(a^2-1))(a^2+1)]^2
=(a^4-1)^2
=a^8-2a^4+1

4) (x-2y-1)(2y-x-1)
=[-1+(x-2y)][-1-(x-2y)]
=1-(x-2y)^2
=1-x^2+4xy-4y^2

Answer 5

1） [(X+3)^2+(x+3)(x-3)]/2x
=(x^2+6x+9+x^2-9)/2x
=x+3
2) (x+1)^2+2(1-x)-x^2
=x^2+2x+1+2-2x-x^2
=3
3) (a-1)^2(a-1)^2(a^2+1)^2
=(a^2-2a+1)(a^2-2a+1)(a^4+2a^2+1)
=(a^4-2a^3+a^2-2a^3+4a^2-2a+a^2-2a+1)(a^4+2a^2+1)
=(a^4-4a^3+6a^2-4a+1)(a^4+2a^2+1)

4) (x-2y-1)(2y-x-1)
=-[(2y-x)^2 - 1]
=-4y^2+4xy-x^2+1