(1+x^2)y'-2xy=(1+x^2)^2求微分的通解
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(1+x^2)y''-2xy=(1+x^2)^2求微分的通解(1+x^2)y''-2xy=(1+x^2)^2求微分的通解(1+x^2)y''-2xy=(1+x^2)^2求微分的通解y=(x^2+1)(x+C
(1+x^2)y'-2xy=(1+x^2)^2求微分的通解
(1+x^2)y'-2xy=(1+x^2)^2求微分的通解
(1+x^2)y'-2xy=(1+x^2)^2求微分的通解
y=(x^2+1)(x+C)
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.(x+y-2)(x+y-2xy)+(xy-1)^2
已知x+y=1,xy=-2,则x²y+xy²-2xy
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(x+y)(x+y+2xy)+(xy+1)(xy-1)(x+y)(x-y)+4(y-1)
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x-y=1,xy=2.求:(-2xy+2x+3y)-(3xy+2y-2x)-(x+4y+xy)的值
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若x-y=4,xy=1,则(2x+y-2xy)-(3x+2y+xy)-(4xy+3y-5x)=