已知|xy-4|+(x-2y-2)²=0
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已知|xy-4|+(x-2y-2)²=0
⑴(-½xy)(x²-4xy+4y²)的值
⑵求(x+2y)²的值
⑴(-½xy)(x²-4xy+4y²)的值
⑵求(x+2y)²的值
![已知|xy-4|+(x-2y-2)²=0](/uploads/image/z/6847015-31-5.jpg?t=%E5%B7%B2%E7%9F%A5%7Cxy-4%7C%2B%EF%BC%88x-2y-2%EF%BC%89%26%23178%3B%3D0)
因为|xy-4|>=0,(x-2y-2)²>=0
所以由已知条件可得:、
xy-4=0
x-2y-2=0
故
xy=4
x-2y=2
从而可知
(-1/2xy)x²-4xy+4y²
=-2(x-2y)²
=-2*2²
=-8
(x+2y)²
=(x-2y)²+8xy
=4+8*4
=36
所以由已知条件可得:、
xy-4=0
x-2y-2=0
故
xy=4
x-2y=2
从而可知
(-1/2xy)x²-4xy+4y²
=-2(x-2y)²
=-2*2²
=-8
(x+2y)²
=(x-2y)²+8xy
=4+8*4
=36
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