已知y₁=5+x,y₂=﹣2x+2,当x取何值时,y₁﹥y₂?-搜答案-神马作文网

已知y/x=2,则求代数式x+y/4x-5y的值分子分母同除以xy；=(1+y/x)/(4-5y/x)=(1+2)/(4-5×2)=3/(-6)=-1/2;您好,很高兴为您解答,skyhunter00

x²+y²-2x+4y+5=0（x²-2x+1)+(y²+4y+4)=0(x-1)²+(y+2)²=0x-1=0x=1y+2=0y=-2x+

y=3x/5原式=x/(x+3x/5)+(3x/5)/[x-3x/5]-(9x^3/25)/(x^3-9x^3/25)=8/3-3/2-9/16=29/48

你的题是什么意思?“^/”是什么意思?

x^2-y^2=(x-y)(x+y)=5×6=30

y=4x/5所以原式=(3x+4x/5)/(x-8x/5)=(19x/5)/(-3x/5)=-19/3

∵x²+y²-4x+2y+5=0x²-4x+4+y²+2y+1=0(x-2)²+(y+1)²=0x-2=0y+1=0∴x=2y=-14x&#

X+Y分之X-Y等于3x=-2yX+Y分之2(x-y)减X+Y分之3X+Y=(-x-3y)/(x+y)=1

因为(x-y)/(x+y)=3,则(x＋y)/(x－y)＝1/3则5(x-y)(x+y)-(x+y)/2(x-y)＝5*3-1/(3*2)=15-1/6=89/6

X^2表示平方X^2+4X+4+Y^2-2Y+1=0(X+2)^2+(Y-1)^2=0因为平方大于=0所以X+2=0Y-1=0X=-2Y=1X^2+Y^2=5

4x=5y,x/y=5/4(x+y)/y=x/y+1=5/4+1=9/4

已知x^2+4y^2-4x+4y+5=0求(（y^4-x^4）/(y-2x)(x+y))*((2x-y)/(xy-y^2))/((x^2+y^2)/y)的值答案：x²+4y²-4x

x²+4x+y²-2y+5=0x²+4x+4+y²-2y+1=0(x+2)²+(y-1)²=0所以x+2=0y-1=0则x=-2y=1所以x

3x-2y5=x+y3交叉相乘,乘积相等：5(x+y)=3(3x-2y)5x+5y=9x-6y11y=4x4x=11y那么x：y=11：4

已知x²+y²+5=2x+4y所以(x-1)²+(y-2)²=0故x=1,y=2所以（2x²-（x+y)(x-y))×((x+y-1)(x-y+1)+

4x=9yx=9/4*y(1)(x+y)/y=[(9/4)y+y]/y=(9/4+1)y/y=9/4+1=13/4(2)(y-x)/2x=[y-(9/4)y]/[2*(9/4)y]=(1-9/4)y/

1）x-5y/x+y=2y-5y/2y+y=-12)x=3z/5,y=4z/52x²-3xy+2xz/2y²-4x²+z²=（18z²/25-36z&

1,-3再问：过程。。。再答：★（x²－2x)+（y²－4y）=5★（x－1）²+（y－2）²=1+4－5★（x－l）²=0,（y－2）²=

原式=x-x+x-x+……-x+(2-1+4-3+5-4+……+2008-2007-2009）y=0+(1×1004-2009)y=-1005y=1005/2