已知2x=3,4y=5,则2x-2y的值为
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平方和绝对值都大于等于0相加等于0,若有一个大于0,则另一个小于0,不成立.所以两个都等于0所以2x+3y-18=0(1)4x+5y-32=0(2)(2)×9-(1)×1636x+45y-288-32
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
(5x+4y)*2=12a(2x+3y)*5=5bb-a:15y-8y=-7y=-1将y=-1带入2x+3y=1,得x=2所以x+y=1
(3x+2y-20)^2+/4x-5y-19/=0由非负性3x+2y-20=0.(1)4x-5y-19=0.(2)(1)*5+(2)*2得23x-138=0x=6代入(1)得18+2y-20=0y=1
y=4x/5所以原式=(3x+4x/5)/(x-8x/5)=(19x/5)/(-3x/5)=-19/3
4x-5y+2z=0(1)x+2y=3z(2)(2)×4-(1)得:13y=14zy=14/13z(1)×2+(2)×5得:13x=11zx=11/13z所以:x:y:z=11/13:14/13:1=
4x+y=5A3x+2y=4BA-B=X-Y=1
x+1/x=3,x^2+1/x^2+2=9x^2+1/x^2=7(x-1/x)^2=X^2+1/x^2-2=7-2=5x-1/x=±√5x^2+y^2-4x-2y+5=0(x-2)^2+(y-1)^2
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
设a(2x+5y+4z)+b(7x+y+3z)=x+y+z比较系数得2a+7b=5a+b=4a+3b=1a=1/11,b=2/11因此x+y+z=a(2x+5y+4z)+b(7x+y+3z)=1/11
平方和绝对值都大于等于0相加等于0,若有一个大于0,则另一个小于0,不成立.所以两个都等于0所以2x+3y-18=0(1)4x+5y-32=0(2)(2)×9-(1)×1636x+45y-288-32
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
2y=x-3带入[(3x+2y)(3x-2y)-(x+2y)(5x-2y)]/4x=[(3x+x-3)(3X-x+3)-(x+x-3)(5x+3-x)]/4x=(8x2-6x=12x-9-8x2=12
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
原式=(9x²-4y²-5x²-8xy+4y²)÷4x=(4x²-8xy)÷4x=x-2y=2011
6x-4y-2(x-5y)=6x-4y-2x+10y=4x+6y=2(2x+3y);∵2x+3y=5,∴原式=2×5=10.故填10.
x-2y/2x+3y=3/44(x-2y)=3(2x+3y)4x-8y=6x+9y-2x=17yx/y=-17/2
2x+y+3=xy,①若x=1,则y不存在,此时5x+4y无最值;②若x≠1,则y=(2x+3)/(x-1),M=5x+4y=5x+(8x+12)/(x-1)=(5x²+3x+12)/(x-
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