要使分式(x 1)×y x^2-y^2的值为零,求x.y的取值范围
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∵x2+y2-2x-4y+5=0,∴x2-2x+1+y2-4y+4=0,(x-1)2+(y-2)2=0,∴x=1,y=2,∴yx−xy=2-12=1.5;故答案为:1.5.
依相反数的意义有|x-y+3|=-|x+y-1999|.因为任何一个实数的绝对值是非负数,所以必有|x-y+3|=0且|x+y-1999|=0.即x−y+3=0①x+y−1999=0②,由①有x-y=
∵3x-5y=0,∴x=5y3,∴原式=5y3−2y5y3+3y=-111.
令y=ux则x^2(xdu+udx)/dx=2(ux)^2+ux^2约掉x^2(xdu+udx)/dx=2(u)^2+u所以(xdu)/dx=2(u)^2之后你该知道了吧求出u关于x的表达式再有y=u
∵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.
2x+yx2-2xy+y2•(x-y)=2x+y(x-y)2•(x-y)(2分)=2x+yx-y;(4分)当x-3y=0时,x=3y;(6分)原式=6y+y3y-y=7y2y=72.(8分)
4xy加5x²y加yx²再问:谢谢了,
x=6-3y &nbs
根据题意,x>0,y>0,则2yx>0,8xy>0,则2yx+8xy≥22yx•8xy=8,即2yx+8xy的最小值为8,若2yx+8xy>m2+2m恒成立,必有m2+2m<8恒成立,m2+2m<8⇔
满足约束条件的平面区域如下图所示:联立x=yx+2y=3可得x=1y=1.即A(1,1)由图可知:当过点A(1,1)时,2x-y取最大值1.故答案为:1
=xy-3xy+2xy-xy=-xy
-x^2y+5xy^2-yx^2=-2x^2y+5xy^2
2xy-y+y-xy=xy
x+2yx2−y2•(x+y)=x+2y(x+y)(x−y)•(x+y)=x+2yx−y;当2x+y=0时,y=-2x,原式=x−4xx+2x=−3x3x=-1.
3x^2y-3xy^2+6yx^2-9y^2x=3x^2y+6yx^2-3xy^2-9y^2x=9x^2y-12xy^2=3xy(3x-4y)
由1x-1y=3,得y-x=3xy,∴x-y=-3xy,∴2x+3xy-2yx-2xy-y=2(x-y)+3xy(x-y)-2xy=-6xy+3xy-3xy-2xy=-3xy-5xy=35.故选:C.
小王上街买苹果和梨一共买了60个,并且苹果的个数是梨的两倍请问小王买了苹果和梨各多少个?
∵x2-4xy+4y2=0,∴(x-2y)2=0,∴x=2y,∴x-yx+y=2y-y2y+y=13.故分式x-yx+y的值等于13.
∵x-y=4xy,∴2x+3xy-2yx-2xy-y=2(x-y)+3xyx-y-2xy=8xy+3xy4xy-2xy=112.故答案为:112.
化简:[(y²-x²)/(5x²-4yx)]/[(x+y)/(5x-4y)]原式=[(y+x)(y-x)/x(5x-4y)]×[(x+y)/(5x-4y)]=(y-x)/