x平方y平方xy3y30,求x的y次方的值
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x/y=3x=3y(x^2+y^2-xy)/(x^2-y^2)=(9y^2+y^2-3y^2)/(9y^2-y^2)=7y^2/8y^2=7/8
两个y²?应该是(2x+y)(4x²-2xy+y²)=2x(4x²-2xy+y²)+y(4x²-2xy+y²)=8x³-
题目肯定有问题,不用纠结了.条件x(y-1)-y(x-1)=4可等价的写为y-x=4.但是对于满足y-x=4的x,y,x²+y²可以有无穷多种取值.例如x=1,y=5满足条件,相应
x^2+y^2=6y-4x-13x^2+4x+4+y^2-6y+9=0(x+2)^2+(y-3)^2=0x=-2y=3x^2-y^2=(-2)^2-3^2=4-9=-5
(x-y)/(x+y)=(x-y)(x+y)/[(x+y)^2]=(x^2-y^2)/[x^2+y^2+2xy]=2xy/[x^2+y^2+x^2-y^2]=2xy/(2x^2)=y/xx^2-y^2
x²+y²-4x+1=0;(x-2)²+y²=3;所以半径r=√3,圆心:(2,0)1)设k=y/x;y=kx;代入原来方程,整理:(1+k²)x&s
原方程左边=(2x-y)(x+2y)=0所以x=y/2或x=-y/2将x的值分别代入到x+y/x-y得-3或-1/3
(x平方+y平方)平方-y平方=x平方+6所以(x²+y²)²-(x²+y²)-6=0(x²+y²+2)(x²+y
平方差原式=(x²-4y²)(x²-4y²)=(x²-4y²)²=x^4-8x²y²+16y^4再问:麻烦你能
x²-4x+y²+6y+13=0﹙x²-4x+4﹚+﹙y²+6y+9﹚=0﹙x-2﹚²+﹙y+3﹚³=0x-2=0,x=2y+3=0,y=﹣
∵/x+y-7/(绝对值)+(xy-5)平方=0∴x+y-7=0xy-5=0∴x+y=7xy=5∴x²+y²=(x+y)²-2xy=7²-2×5=39
y=2x/3则xy=2x2/3y2=4x2/9所以原式=(2x2/3)/(x2+4x2/9)-(4x2/9)/(x2-4x2/9)=(2/3)/(1+4/9)-(4/9)/(1-4/9)=6/13+4
因为x²+y²-4x+2y+5=0所以(x²-4x+4)+(y²+2y+1)=0所以(x-2)²+(y+1)²=o所以x=2y=-1所以下面
X+Y=√(X平方+2XY+Y平方)
将xy=8代入x平方y-8xy平方-x+y=56得8x-8y-x+y=56x-y=8x平方+y平方=(x-y)^2+2xy=8^2+2*8=64+16=80
x{x-1}-{x^2-y}=-3即x^2-x-x^2+y=-3-x+y=-3x-y=3而x^2+y^2-2xy=(x-y)^2=3^2=9
x平方-y平方=4(x+y)(x-y)=4(x+y)平方*(x-y)平方=4*4=16
考点:换元法解一元二次方程分析:设x2+y2=m,化简方程后求得m的值即可.设x2+y2=m,方程化为(m+2)(m-1)=0∴m1=-2,m2=1∵x2+y2≥0∴m1=-2舍去,即x2+y2=1.
=(x²+y²)²-(2xy)²=(x²+2xy+y²)(x²-2xy+y²)=(x+y)²(x-y)
根据:a²-b²=(a+b)(a-b)(x²+y²)²-(2xy)²假设:(x²+y²)²=a²,(