化简1.(3y-2y+z)(3y-2y-z)2.(2a+b-c+3d)(2a-b+c+3d)计算(1+1/2)[1+1/(2^2)][1+1/(2^4)][1+1/(2^8)]+1/(2^15)化简求值(x-y)^2*(x^2+xy+y^2)^2-(x^3+y^3)*(-x^3+y^3)其中x=1,y=-1不好意思,第一题确实打错了。是(3x-2y+z)(3x-2y-z)但不能用

来源:学生作业帮助网 编辑:作业帮 时间:2024/11/17 10:42:48

化简1.(3y-2y+z)(3y-2y-z)2.(2a+b-c+3d)(2a-b+c+3d)计算(1+1/2)[1+1/(2^2)][1+1/(2^4)][1+1/(2^8)]+1/(2^15)化简求值(x-y)^2*(x^2+xy+y^2)^2-(x^3+y^3)*(-x^3+y^3)其中x=1,y=-1不好意思,第一题确实打错了。是(3x-2y+z)(3x-2y-z)但不能用
化简
1.(3y-2y+z)(3y-2y-z)
2.(2a+b-c+3d)(2a-b+c+3d)
计算
(1+1/2)[1+1/(2^2)][1+1/(2^4)][1+1/(2^8)]+1/(2^15)
化简求值
(x-y)^2*(x^2+xy+y^2)^2-(x^3+y^3)*(-x^3+y^3)
其中x=1,y=-1
不好意思,第一题确实打错了。
是(3x-2y+z)(3x-2y-z)
但不能用设元法,用公式来计算。

化简1.(3y-2y+z)(3y-2y-z)2.(2a+b-c+3d)(2a-b+c+3d)计算(1+1/2)[1+1/(2^2)][1+1/(2^4)][1+1/(2^8)]+1/(2^15)化简求值(x-y)^2*(x^2+xy+y^2)^2-(x^3+y^3)*(-x^3+y^3)其中x=1,y=-1不好意思,第一题确实打错了。是(3x-2y+z)(3x-2y-z)但不能用
你的题目应该写错了.我按我的理解应该是:
1、(3x-2y+z)(3x-2y-z)
设3x-2y=a
原式=(a+z)(a-z)=a^2 - z^2=(3x-2y)^2-z^2=9x^2-12xy+4y^2-z^2
2、设2a+3d=x b-c=y
原式=(x+y)(x-y)=x^2-y^2=(2a+3d)^2-(b-c)^2=4a^2+12ad+9d^2-b^2+2bc-c^2
3、原式=(1-1/2)(1+1/2)〔1+1/(2^2)〕〔1+1/(2^4)〕〔1+1/(2^8)〕÷(1-1/2)+1/(2^15)=(1-2^16)×2+1/(2^15)=2
4、原式=(x-y)(x^2+xy+y^2)〔(x-y)(x^2+xy+y^2)+(x+y)(x^2-xy+y^2)〕=(x^3-y^3)(x^3-y^3+x^3+y^3)=2x^3(x^3-y^3)=2*1*2=4

化简
1.(3y-2y+z)(3y-2y-z)
=(y+z)(y-z)
=y^2-z^2
2.(2a+b-c+3d)(2a-b+c+3d)
=(2a+3d+b-c)(2a+3d-b+c)
=(2a+3d)^2-(b-c)^2
=4a^2+12ad+9d^2-b^2+2bc-c^2
计算
(1+1/2)[1+1/(2^2)][1...

全部展开

化简
1.(3y-2y+z)(3y-2y-z)
=(y+z)(y-z)
=y^2-z^2
2.(2a+b-c+3d)(2a-b+c+3d)
=(2a+3d+b-c)(2a+3d-b+c)
=(2a+3d)^2-(b-c)^2
=4a^2+12ad+9d^2-b^2+2bc-c^2
计算
(1+1/2)[1+1/(2^2)][1+1/(2^4)][1+1/(2^8)]+1/(2^15)
=(1-1/2)(1+1/2)[1+1/(2^2)][1+1/(2^4)][1+1/(2^8)]÷(1-1/2)+1/(2^15)
=(1-1/2^2)[1+1/(2^2)][1+1/(2^4)][1+1/(2^8)]÷(1-1/2)+1/(2^15)
=[1-1/(2^4)][1+1/(2^4)][1+1/(2^8)]÷(1-1/2)+1/(2^15)
=[1-1/(2^8)][1+1/(2^8)]÷(1-1/2)+1/(2^15)
=[1-1/(2^16)]÷(1-1/2)+1/(2^15)
=2-1/(2^15)+1/(2^15)
=2
化简求值
(x-y)^2*(x^2+xy+y^2)^2-(x^3+y^3)*(-x^3+y^3)
其中x=1,y=-1
原式=(x^3-y^3)^2-(x^3+y^3)*(-x^3+y^3)
=(x^3-y^3)[(x^3-y^3)+(x^3+y^3)]
=(x^3-y^3)2x^3
=2x^6-2(xy)^3
=2+2
=4

收起