y 2=2x-1 2y=x-3
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原式=[2(2x+y)]2-3x2-y2+2-2y2=4(2x+y)2-3(x2+y2)+2∵2x+y=7,x2+y2=5∴原式=183.
∵y1与3x成正比例y2与(x+5)成正比例∵y1=k1×3xy2=k2(x+5)(k1k2≠0)∵y=2y1-y2=6k1-k2(x+5)∴12=6k1-6k2-2=-6k1-4k2∴k1=1k2=
[(x+y)2-(x+y)(x+3y)-4y2]÷2y,=[x2+2xy+y2-x2-3xy-xy-3y2-4y2]÷2y=(-2xy-6y2)÷2y=-x-3y,当x=-2,y=12时,原式=-(-
即(x²-6x+9)+(y²+4y+4)=0(x-3)²+(y+2)²=0所以x-3=y+2=0所以x=3,y=-2所以2x+3y=6-6=0
已知2x=3y,求xy/(x^2+y^2)-y^2/(x^2-y^2)的值2x=3y-->x=(3/2)yx^2=(9/4)y^2xy/(x^2+y^2)-y^2/(x^2-y^2)==(3/2)y*
y^2=x^3-3x^2+2xx^2=y^3-3y^2+2y两式相减得:y^2-x^2=(x^3-y^3)-3(x^2-y^2)+2(x-y)(x-y)(x^2+xy+y^2-2x-2y+2)=0所以
原式=xy3+xy2-xy3-2xy2-3xy=-xy2-3xy,由x+y=4,x-y=6可知:x=5,y=-1,把x=5,y=-1代入,原式=10.
∵3x2+2y2=6x,∴y2=-32x2+3x,由y2=-32x2+3x≥0,可得0≤x≤2,又x2+y2=x2-32x2+3x=-12x2+3x=-12(x-3)2+92,∵0≤x≤2,∴x=2时
[(x+2y)2-(x+y)(3x-y)-5y2]÷2x=(x2+4xy+4y2-3x2-2xy+y2-5y2)÷2x=(-2x2+2xy)÷2x=y-x,当x=-2,y=12时,原式=12-(-2)
分解因式有(x-3y)(2x-y)=0所以有x=3y或2x=y所以x:y=3:1或x:y=1:2
(X+Y)²=X²+Y²+2XY=X²+Y²+X²-Y²=2X²(X-Y)²=X²+Y²-
解x+y=8两边平方(x+y)²=64即x²+2xy+y²=64∵xy=12∴x²+y²+24=64∴x²+y²=40(x-y)&
x²+y²+2x-2√3y=0,即(x+1)²+(y-√3)²=4=2²,是一个以(-1,√3)为圆心,2为半径的圆,且过原点因此(1)x²
x²-2xy+y²/x²-y²=(x-y)²/(x-y)(x+y)=(x-y)/(x+y)因为x=3,y=-5,所以(3-(-5))/(3+(-5))
Y2>Y3>Y1a=0ora^2+2a+1-4a=0a=1
x2+3y2+6x-12y+21=0(x²+6x+9)+3(y²-4y+4)=0(x+3)²+3(y-2)²=0平方大于等于0,相加等于0,若有一个大于0,则另
y=2x-3y2=2x2-3y-y2=(2x-3)-(2x2-3)=2(x-x2)x再问:应该是这样的、、有逗号。。(x,y),(x2,y2)再答:额。。不影响结果。。
解题思路:运用完全平方公式和平方差公式进行化简,再代入数值进行计算解题过程:解:
(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