y=1 2x*x与x*x y*y=8所围成图像的面积
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x/(xy-y²)=x/[y(x-y)]y/(x²-xy)=y/[x(x-y)]∴公分母为:xy(x-y)通分后得x/(xy-y²)=x²/[xy(x-y)]y
(x+y)(x²-xy+y²)=x³-x²y+xy²+x²y-xy²+y³=x³+y³(x-y)(x
原式=(x-y)(x²+2xy+y²-xy)-(x-y)(x²-2xy+y²+xy)=(x-y)(x²+xy+y²)-(x+y)(x&sup
1.相等y=6/xx^2+(6/x)^2=12解得x^2=6,y^2=6xy=6x=y2.x=3y=3z=0x=6-y(6-y)y=z^2+96y-y^2=z^2+9y^2-6y+9+z^2=0(y-
设根号下x+y=a,则a平方+a=12可算出a1=-4,舍去a2=3所以x+y=9再和x+xy+y=23组成方程组,解得x1=7,y1=2x2=7y2=7
x^2+xy+y=14y^2+xy+x=28两式相加x^2+y^2+2xy+x+y=42(x+y)^2+(x+y)-42=0(x+y-6)(x+y+7)=0x+y=6或x+y=-7
如果|x|/x+|y|/y=0,试比较-x/y与xy的大小.因为有-x/y首先x,y不等于0!所以|x|/x和|y|/y互为相反数!所以|y|=|x|,y=-x-x/y=1xy=-1所以-x/y>xy
(x-y)[(x+y)(x+y)-xy]+(x+y)[(x-y)(x-y)+xy]=(x-y)(x+y)^2-(x-y)xy+(x+y)(x-y)^2+(x+y)xy=(x-y)(x+y)(x+y+x
x(x-y)-y(y-x)=12那么:化简得:(x+y)*(x—y)=12x、y是自然数,所以x1=2,y1=4x2=4,y2=2所以x+y-xy=-2
假设x²-xy=a相加x²-xy+xy-y²=a-12所以x²-y²=a-12相减x²-xy-xy+y²=a+12所以x²
你确定没写错题目?后面的式子一约就变成前面的式子了,答案也是8/xy再问:怎么个约法?再答:y/xyyy约掉剩下1/x
(4xy+12y)+[7x-(3xy+4y-x)]=4xy+12y+7x-3xy-4y+x=xy+8x+8y=xy+8(x+y)=(-2)+8*3=-2+24=22
即(x-2y)²=0x-2y=0所以x=2y所以原式=(2x²+2xy-xy-y²)/(4x²-4xy+y²)=(2x²+xy-y²
7或者-8再问:求过程^_^再答:两个等式两边相加
第一个式子上下同乘x-y得2x平方y-2xy平方/(x+y)(x平方-y平方)第二个式子上下同乘x+y得x平方y/(x+y)(x平方-y平方)分母一样了
令y=kxx*x+kx*x=k*x+k*k*x*x(1-k*k+k)x^2-kx=0x((1-k*k+k)x-k)=0由上式得X=0或(1-k*k+k)x-k=0解得:k=(x-1+(或-)√((1-
(x+y)(x+y)=25x^2+2xy+y^2=25……(1)(x-y)(x-y)=9x^2-2xy+y^2=9……(2)(1)+(2)得:2x^2+2y^2=34x^2+y^2=17(1)-(2)
2边同时除以y*y,得到12(x/y)^2+12=25(x/y).然后算出就可以了.
xy=1,x与y成(反)比例,x=y/5,x与y成(反)比例,x/3=y/4,x与y成(正)比例,3/x=y,x与y成(反)比例
y=4-2x"还有做法"6x^2+3xy-12x-9=3x(x+y-4)-9(x+y-4)=0时,6x^2+3xy-12x-9的值与x无关.--〉y=4-2x