1.(3+1)(3^2+1)(3^4+1)(3^8+1)(3^16+1)
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1.(3+1)(3^2+1)(3^4+1)(3^8+1)(3^16+1)
2.已知x(x+1)-(x^2+y)=-3,求[(x^2+y^2)/2]-xy的值
3.(x+y+z)(x-y-z)=?
特别是第三题以及此类题如何运用公式.
2.已知x(x+1)-(x^2+y)=-3,求[(x^2+y^2)/2]-xy的值
3.(x+y+z)(x-y-z)=?
特别是第三题以及此类题如何运用公式.
![1.(3+1)(3^2+1)(3^4+1)(3^8+1)(3^16+1)](/uploads/image/z/1196111-47-1.jpg?t=1.%EF%BC%883%2B1%EF%BC%89%EF%BC%883%5E2%2B1%EF%BC%89%EF%BC%883%5E4%2B1%EF%BC%89%EF%BC%883%5E8%2B1%EF%BC%89%EF%BC%883%5E16%2B1%EF%BC%89)
1、(3+1)(3^2+1)(3^4+1)(3^8+1)(3^16+1)
=(3+1)(3^2+1)(3^4+1)(3^8+1)(3^16+1)*(3-1)/(3-1)
=(3^2-1)(3^2+1)(3^4+1)(3^8+1)(3^16+1)/(3-1)
=(3^4-1)(3^4+1)(3^8+1)(3^16+1)/(3-1)
=(3^8-1)(3^8+1)(3^16+1)/(3-1)
=(3^16-1)(3^16+1)/(3-1)
=(3^32-1)/2
2、x(x+1)-(x^2+y)=-3,整理得:
x^2+x-x^2-y=-3
x-y=-3
[(x^2+y^2)/2]-xy=(x^2-2xy+y^2)/2
=(x-y)^2/2
=9/2
3、(x+y+z)(x-y-z)=[x+(y+z)][x-(y+z)]
=x^2-(y+z)^2
=x^2-y^2-z^2-2yz
=(3+1)(3^2+1)(3^4+1)(3^8+1)(3^16+1)*(3-1)/(3-1)
=(3^2-1)(3^2+1)(3^4+1)(3^8+1)(3^16+1)/(3-1)
=(3^4-1)(3^4+1)(3^8+1)(3^16+1)/(3-1)
=(3^8-1)(3^8+1)(3^16+1)/(3-1)
=(3^16-1)(3^16+1)/(3-1)
=(3^32-1)/2
2、x(x+1)-(x^2+y)=-3,整理得:
x^2+x-x^2-y=-3
x-y=-3
[(x^2+y^2)/2]-xy=(x^2-2xy+y^2)/2
=(x-y)^2/2
=9/2
3、(x+y+z)(x-y-z)=[x+(y+z)][x-(y+z)]
=x^2-(y+z)^2
=x^2-y^2-z^2-2yz
1.(3+1)(3^2+1)(3^4+1)(3^8+1)(3^16+1)
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