1.(x+2y)(x-y)=0, x=-2y或x=y.
若x=-2y
则原式=(4y2-6y2+y2)/(4y2+y2)
=-y2/5y2
=-1/5
若x=y
则原式=(y2+3y2+y2)/(y2+y2)
=5y2/2y2
=5/2
故多项式=-1/5或5/2
2.设根号-a=x,根号-b=y
则条件等式为 x-y2-2xy=y-x
故 2x-2xy=y2+y
2x(1-y)=y(y+1)
x=y(y+1)/(1-y)
(x+2y)(x-y)=0
x=-2y或x=y
x=-2y
原式=(4y²-6y²+y²)/(4y²+y²)
=-y²/5y²
=-1/5
x=y
原式=(y²+3y²+y²)/(y²+y²)
=5y²/2y²
=5/2
所以原式=-1/5或5/2
第二题目看不清楚啊. 加上括号.
解:1.(x+2y)(x-y)=0
x=-2y或x=y
若x=-2y
则原式=(4y²-6y²+y²)/(4y²+y²)
=-y²/5y²
=-1/5
若x=y
则原式=(y²+3y²+y²)/(y²+y²)
=5y²/2y²
=5/2
故多项式=-1/5或5/2
2.设根号-a=x,根号-b=y
则条件等式为 x-y²-2xy=y-x
故 2x-2xy=y²+y
2x(1-y)=y(y+1)
x=y(y+1)/(1-y)
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