Yx^4-8X^2 2
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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.
方程x2+y2-4x+1=0表示以点(2,0)为圆心,以3为半径的圆.设yx=k,即y=kx,由圆心(2,0)到y=kx的距离为半径时直线与圆相切,斜率取得最大、最小值,由|2k−0|k2+1=3,解
yx+xy=xyx+xyy=xy(x+y)xy,∵x+y=8,xy=6,∴原式=6×86=463.
xy+xz=8-x²yx+yz=12-y²zy+zx=-4-z²x(x+y+z)=8y(x+y+z)=12z(x+y+z)=-4(x+y+z)²=8+12-4=
∵x2+y2-6x-8y+25=0,∴(x-3)2+(y-4)2=0,∴x=3,y=4,当x=3,y=4时,原式=43-34=712.
∵x+y=4,xy=3,∴原式=x2+y2xy=(x+y)2−2xyxy=16−63=103.
此方程有无数解这里要把x^2+3x看做一个整体
∵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.
以x为主元,将方程整理为3x2-(3y+7)x+(3y2-7y)=0,∵x是整数,∴△=[-(3y+7)]2-4×3(3y2-7y)≥0,∴21−1439≤y≤21+1439,∴整数y=0,1,2,3
∵1-8x≥0,8x-1≥0,∴x=18,y=12,∴代数式xy+yx+2-xy+yx−2=14+4+2-14+4−2=52-32=1.故选:B.
解析:在有理数范围里分解-3xy²-8yx-2x=-x(3y^2+8y+2)在实数范围里分解-3xy²-8yx-2x=-x(3y^2+8y+2)=-3x[(y-4/3)^2-10/
原式=[(x+y)2(x-y)(x+y)+-4xy(x-y)(x+y)]×(x+3y)(x-3y)(x+3y)(x-y)=x-3yx+y,由已知得(3x-2y)(x+y)=0,因为x+y≠0,所以3x
似乎题目应该是y=√(x-2)+√(2-x)+4x-2>=02-x>=0x=2代入得y=4yx=4*2=8y的x次=4²=16
(1)x=4代入,4+y=4y-2y;所以y=4(2)y=4代入,4+4=4x-2*4;所以x=4
-x^2y+4x^2y-3yx^2-7x^2y^2+|-8x^2y^2|=-x²y+4x²y-3x²y-7x²y²+8x²y²=(
∵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.
x²y²-8xy+4x²+y²+4=0x²y²-4xy+4+4x²-4xy+y²=0(xy-2)²+(2x-y
解答如下:x+2y=(yx)/44x+8y=xyxy-8y=4x(x-8)y=4x当x≠8时(x=8不成立)y=4x/(x-8)x+2y=(2x+1)/32y=(2x+1)/3-x2y=(1-x)/3
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)/