已知xy=8满足s^2*y-8x*y^2-x+y=56,求x^2+y^2的值(答案为80)1.若x-3=y-2=z-1,求x^2+y^2+z^2-xy-yz-zx的值(答案为3)2.对于式子x^2+y^2-x+6y+41/4,你能否确定其值的正负性?若能,;若不能,请简要说明理由.
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![已知xy=8满足s^2*y-8x*y^2-x+y=56,求x^2+y^2的值(答案为80)1.若x-3=y-2=z-1,求x^2+y^2+z^2-xy-yz-zx的值(答案为3)2.对于式子x^2+y^2-x+6y+41/4,你能否确定其值的正负性?若能,;若不能,请简要说明理由.](/uploads/image/z/5480783-71-3.jpg?t=%E5%B7%B2%E7%9F%A5xy%3D8%E6%BB%A1%E8%B6%B3s%5E2%2Ay-8x%2Ay%5E2-x%2By%3D56%2C%E6%B1%82x%5E2%2By%5E2%E7%9A%84%E5%80%BC%EF%BC%88%E7%AD%94%E6%A1%88%E4%B8%BA80%EF%BC%891.%E8%8B%A5x-3%3Dy-2%3Dz-1%2C%E6%B1%82x%5E2%2By%5E2%2Bz%5E2-xy-yz-zx%E7%9A%84%E5%80%BC%EF%BC%88%E7%AD%94%E6%A1%88%E4%B8%BA3%EF%BC%892.%E5%AF%B9%E4%BA%8E%E5%BC%8F%E5%AD%90x%5E2%2By%5E2-x%2B6y%2B41%2F4%2C%E4%BD%A0%E8%83%BD%E5%90%A6%E7%A1%AE%E5%AE%9A%E5%85%B6%E5%80%BC%E7%9A%84%E6%AD%A3%E8%B4%9F%E6%80%A7%3F%E8%8B%A5%E8%83%BD%2C%EF%BC%9B%E8%8B%A5%E4%B8%8D%E8%83%BD%2C%E8%AF%B7%E7%AE%80%E8%A6%81%E8%AF%B4%E6%98%8E%E7%90%86%E7%94%B1%EF%BC%8E)
已知xy=8满足s^2*y-8x*y^2-x+y=56,求x^2+y^2的值(答案为80)1.若x-3=y-2=z-1,求x^2+y^2+z^2-xy-yz-zx的值(答案为3)2.对于式子x^2+y^2-x+6y+41/4,你能否确定其值的正负性?若能,;若不能,请简要说明理由.
已知xy=8满足s^2*y-8x*y^2-x+y=56,求x^2+y^2的值(答案为80)
1.若x-3=y-2=z-1,求x^2+y^2+z^2-xy-yz-zx的值(答案为3)
2.对于式子x^2+y^2-x+6y+41/4,你能否确定其值的正负性?若能,;若不能,请简要说明理由.
已知xy=8满足s^2*y-8x*y^2-x+y=56,求x^2+y^2的值(答案为80)1.若x-3=y-2=z-1,求x^2+y^2+z^2-xy-yz-zx的值(答案为3)2.对于式子x^2+y^2-x+6y+41/4,你能否确定其值的正负性?若能,;若不能,请简要说明理由.
x^2y-8xy^2-x+y=56
xy(x-y)-(x-y)=56
xy=8
所以8(x-y)-(x-y)=56
7(x-y)=56
x-y=8
两边平方
x^2-2xy+y^2=64
x^2+y^2=64+2xy=64+2*8=80
x-3=y-2
所以x-y=1
x-3=z-1
所以x-z=2
y-2=z-1
所以y-z=1
x^2+y^2+z^2-xy-yz-zx
=(2x^2+2y^2+2z^2-2xy-2yz-2zx)/2
=[(x^2-2xy+y^2)+(y^2-2yz+z^2)+(x^2-2xz+z^2)]/2
=[(x-y)^2+(y-z)^2+(z-x)^2]/2
=(1^2+1^2+2^2)/2
=3
x^2+y^2-x+6y+41/4
=(x^2-x+1/4)+(y^2+6y+9)+1
=(x-1/2)^2+(y+3)^2+1
完全平方大于等于0
所以(x-1/2)^2+(y+3)^2>=0
所以(x-1/2)^2+(y+3)^2+1>0
所以是正数
x-3=y-2=z-1
x-y=1 y-z=1 x-z=2
(x-y)^2=1 (y-z)^2=1 (x-z)^2=4
x^2+y^2+z^2-xy-yz-zx
=2(x^2+y^2+z^2-xy-yz-zx)/2
=[(x-y)^2+(x-z)^2+(y-z)^2]/2
=[1+1+4]/2
=3
x^2+y^2-x+6y+41/4
=x^2-x+1/4+y^2+6y+9+1
=(x-1/2)^2+(y+3)^2+1
(x-1/2)^2≥0 (y+3)^2≥0
=(x-1/2)^2+(y+3)^2+1≥1