x加y x减y分之y方
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x减2y分之x减y除以x方减4xy加4y方分之x方减y方=[(x-y)/(x-2y)]×[(x-2y)²/(x+y)(x-y)]=(x-2y)/(x+y)=[(1+√2-2(1-√2)]/(
y=√(x-2)+√(2-x)+3有意义则需x-2≥0且2-x≥0只有x-2=0,∴x=2∴y=0+0+3=3y^x=3²=9y^x的平方根为±√9=±3
1.y'=sinx+xcosx2.
c再问:真的是c吗?再答:百分百正确再问:那好谢谢你了。
【俊狼猎英】团度为您解答~原式=[(x^2-y^2)*2(x-y)^2]/[xy(xy-y^2)]先约掉一个(x-y),分子分母同时处以y^3得到2[(y/x)^2-1]*[(y/x)-1]/[(y/
∵x−yx+y=2,∴x-y=2(x+y),∴x−yx+y-2x+2yx−y=2(x+y)x+y-2(x+y)2(x+y)=2-1=1,故答案为:1.
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已知:(x-1)(y-2)-x(y-3)=8,那么:xy-2x-y+2-xy+3x=8化简得:x-y=6所以:2分之(x²+y²)-xy=2分之(x²+y²-2
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(x^2+y^2)*(x^2-1+y^2)-12=0可以设x^2+y^2=z那么有z(z-1)=12z^2-z-12=0(z-4)(z+3)=0z=4或z=-3但z是平方和一定大于0所以z=4记得采纳
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原式=[x/(x-y)]*[y^2/(x+y)]-[x^4y/(x^2+y^2)(x^2-y^2)]/[x^2/(x^2+y^2)]=xy^2/(x^2-y^2)-x^2y/(x^2-y^2)=xy(
椭圆半焦距:2a^2-2b^2双曲线半焦距:a^2+b^2有相同焦点,2a^2-2b^2=a^2+b^2a^2=3b^2椭圆半焦距:c^2=2a^2-2b^2=4b^2椭圆半焦距c=2b椭圆半长轴=根
x(x-1)-(x^2-y)=-2,得x-y=2(x^2+y^2)/2-xy=(x^2-2xy+y^2)/2=(x-y)^2/2=2