分解因式:[(2x^2-(x+y)(x-y)][(z-x)(x+z)-(-y-z)(z-y)]+z,其中x=‐1,y=1/2,z=‐3/4
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![分解因式:[(2x^2-(x+y)(x-y)][(z-x)(x+z)-(-y-z)(z-y)]+z,其中x=‐1,y=1/2,z=‐3/4](/uploads/image/z/8769689-17-9.jpg?t=%E5%88%86%E8%A7%A3%E5%9B%A0%E5%BC%8F%3A%5B%282x%5E2-%28x%2By%29%28x-y%29%5D%5B%28z-x%29%28x%2Bz%29-%28-y-z%29%28z-y%29%5D%2Bz%2C%E5%85%B6%E4%B8%ADx%3D%E2%80%901%2Cy%3D1%2F2%2Cz%3D%E2%80%903%2F4)
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分解因式:[(2x^2-(x+y)(x-y)][(z-x)(x+z)-(-y-z)(z-y)]+z,其中x=‐1,y=1/2,z=‐3/4
分解因式:[(2x^2-(x+y)(x-y)][(z-x)(x+z)-(-y-z)(z-y)]+z,其中x=‐1,y=1/2,z=‐3/4
分解因式:[(2x^2-(x+y)(x-y)][(z-x)(x+z)-(-y-z)(z-y)]+z,其中x=‐1,y=1/2,z=‐3/4
:[(2x^2-(x+y)(x-y)][(z-x)(x+z)-(-y-z)(z-y)]+z
=(2x²-x²+y²)(z²-x²-y²+z²)+z
=(x²+y²)[2z²-(x²+y²)]+z
=2z²(x²+y²)-(x²+y²)²+z
当x=‐1,y=1/2,z=‐3/4时,上式得:
2*(-3/4)²[(-1)²+(1/2)²]-[(-1)²+(1/2)²]²+(-3/4)
=2*9/16(1+1/4)-(1+1/4)²-3/4
=9/8*5/4-20/16-3/4
=1/32(45-40-24)
=-19/32
(x^2+y^2)(y^2-x^2)+z=-x^4+y^4+z=5/16
xx=1 yy=1/4 zz=9/16
[2xx-xx+yy] [zz-xx+zz-yy] + z
= (xx+yy)[2zz-(xx+yy)] +z
= 5/4 * (9/8 - 5/4) - 3/4
= -5/32 - 3/4
= -29/32
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