只求第三问已知|xy-4|+(x-2y-2)²=0 (1)求x²-4xy+4y²的值 (2)求(x+2y)²的值 (3)求(x²-4y²)(2x+4y)的值
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![只求第三问已知|xy-4|+(x-2y-2)²=0 (1)求x²-4xy+4y²的值 (2)求(x+2y)²的值 (3)求(x²-4y²)(2x+4y)的值](/uploads/image/z/1611977-41-7.jpg?t=%E5%8F%AA%E6%B1%82%E7%AC%AC%E4%B8%89%E9%97%AE%E5%B7%B2%E7%9F%A5%7Cxy-4%7C%2B%EF%BC%88x-2y-2%EF%BC%89%26%23178%3B%3D0+%EF%BC%881%EF%BC%89%E6%B1%82x%26%23178%3B-4xy%2B4y%26%23178%3B%E7%9A%84%E5%80%BC+%EF%BC%882%EF%BC%89%E6%B1%82%EF%BC%88x%2B2y%EF%BC%89%26%23178%3B%E7%9A%84%E5%80%BC+%EF%BC%883%EF%BC%89%E6%B1%82%EF%BC%88x%26%23178%3B-4y%26%23178%3B%EF%BC%89%EF%BC%882x%2B4y%EF%BC%89%E7%9A%84%E5%80%BC)
只求第三问已知|xy-4|+(x-2y-2)²=0 (1)求x²-4xy+4y²的值 (2)求(x+2y)²的值 (3)求(x²-4y²)(2x+4y)的值
只求第三问
已知|xy-4|+(x-2y-2)²=0 (1)求x²-4xy+4y²的值 (2)求(x+2y)²的值 (3)求(x²-4y²)(2x+4y)的值
只求第三问已知|xy-4|+(x-2y-2)²=0 (1)求x²-4xy+4y²的值 (2)求(x+2y)²的值 (3)求(x²-4y²)(2x+4y)的值
xy-4=0
x-2y-2=0
xy=4
x-2y=2
x²-4xy+4y²
=(x-2y)²
=2²
=4
(x+2y)²
=(x-2y)²+8xy
=4+8*4
=36
(x²-4y²)(2x+4y)
=2(x-2y)(x+2y)²
=2*2*36
=144
|xy-4|+(x-2y-2)²=0
∵|xy-4|≥0,(x-2y-2)²≥0
∴XY-4=0,X-2Y-2=0
∴X=2Y+2
∴XY=(2Y+2)Y=4
∴Y²+Y-2=0
∴(Y+2)(Y-1)=0
∴Y1=1,Y2=-2
∴X1=4/Y1=4,X2=4/Y2=-2
(3)(x²...
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|xy-4|+(x-2y-2)²=0
∵|xy-4|≥0,(x-2y-2)²≥0
∴XY-4=0,X-2Y-2=0
∴X=2Y+2
∴XY=(2Y+2)Y=4
∴Y²+Y-2=0
∴(Y+2)(Y-1)=0
∴Y1=1,Y2=-2
∴X1=4/Y1=4,X2=4/Y2=-2
(3)(x²-4y²)(2x+4y)
=(x-2y)(x+2y)2(x+2y)
=2(x-2y)(x+2y)²
下面的都是你前两问求的
这是我在静心思考后得出的结论,
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|xy-4|+(x-2y-2)²=0
即
xy-4=0,x-2y-2=0
所以
xy=4
x-2y=2
平方,得
(x-2y)²=4
所以
(x+2y)²=(x-2y)²+8xy
=2²+8×4
=36
从而
(x²-4y²)(2x+4y)
=2(x-2y)(x+2y)(x+2y)
=2(x-2y)(x+2y)²
=2×2×36
=144
已知|xy-4|+(x-2y-2)²=0,则
xy-4=0
x-2y-2=0,x-2y=2
(1)x²-4xy+4y²=(x-2y)²=4
(2)(x+2y)²=x²+4xy+4y²=(x-2y)²+8xy=4+8*4=36
(3)(x²-4y²)(2x+4y)=(x-2y)(x+2y)2(x+2y)=2(x-2y)(x+2y)²=36*4=144
xy-4=0
xy=4
x-2y-2=0
x-2y=2
x²-4xy+4y²
=(x-2y)²
=2²
=4
(x+2y)²
=x²+4xy+4y²-4xy+4xy
=(x-2y)²+8xy
=4+8×4
=36
(...
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xy-4=0
xy=4
x-2y-2=0
x-2y=2
x²-4xy+4y²
=(x-2y)²
=2²
=4
(x+2y)²
=x²+4xy+4y²-4xy+4xy
=(x-2y)²+8xy
=4+8×4
=36
(x²-4y²)(2x+4y)
=2(x-2y)(x+2y)(x+2y)
=2×2×36
=144
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