已知|5x-y-8| (7y-x 2)=0则4x 6y

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已知5I(x+9y)(x,y都是整数),求证:5I(8x+7y)

已知5I(x+9y)(x,y都是整数)则,x+9y=5M(M为整数)8x+7y=8(x+9y)-65y=8*5M-65y=5*(8M-13y)因为M和y都是整数,所以,8M-13y为整数咯,假设整数Z

已知(2y-x)/(7x-5y),则x/y的值为( )

(2y-x)/(7x-5y)=2/3即2*(7x-5y)=3*(2y-x)那么14x-10y=6y-3x即17x=16yx/y=16/17

已知(3x+1)(3x-3)-3x+2)^2=-7,(y+7)(y-7)-y(y-7)=0,求8y^2-5y(-y+3x

(3x+1)(3x-3)-(3x+2)^2=9x^2-6x-3-(9x^2+12x+4)=-18x-7=-7∴x=0(y+7)(y-7)-y(y-7)=y^2-49-y^2+7y=7y-49=0∴y=

计算(x^8-x^7y+x^5y^3-x^4y^4+x^3y^5-x^y^7+y^8)(x^2+xy+y^2)

-x^y^7+y^8)(这里有问题,x的几次方?再问:就是一次方再答:(x^8-x^7y+x^5y^3-x^4y^4+x^3y^5-xy^7+y^8)(x^2+xy+y^2)=(x^8-x^7y+x^

已知二元一次方程{3x+y=7 {5x-2y=8,

由题意:①×2﹢②得:11x=22将x=2代入①得:y=1即:x=2y=1

已知x-2y=2012,求[(3x+2y)(3x-2y)-(x+2y)(5x-2y)]/(8x))

[(3x+2y)(3x-2y)-(x+2y)(5x-2y)]/(8x)=[9x²-4y²-(5x²+10xy-2xy-4y²)]/(8x)=(9x²-

已知x-2y=2012,求[(3x+2y)(3x-2y)-(x+2y)(5x-2y)]÷8x的值

[(3x+2y)(3x-2y)-(x+2y)(5x-2y)]÷8x=(9x^2-4y^2-5x^2+2xy-10xy+4y^2)/(8x)=(4x^2-8xy)/(8x)=(x-2y)/2=2012/

已知实数x.y满足根号(x+y-8)+根号(8-x-y)=根号(3x-y-4)+根号(x-2y+7),求x,y

根号(x+y-8)+根号(8-x-y)=根号(3x-y-4)+根号(x-2y+7),根据二次根式有意义得:X+Y-8≥0,8-X-Y≥0,∴X+Y≥8,X+Y≤8,∴X+Y=8,左边为0,右边两个非负

已知x/5=y/3,则(x/x+y)+(y/x-y)-(yxy/xxx-yxy)

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

已知x.x+y.y-6x-8y+25=0,求3x.x+5y的值

可以得出x.x+y.y-6x-8y+25=x.x-6x+9+y.y-8y+16=(x-3).(x-3)+(y-4).(y-4)=0因为(x-3).(x-3)>=0,(y-4).(y-4)>=0所以x=

已知二元一次方程组5x+8y=183x−y=7

5x+8y=18①3x−y=7②,①-②得:2x+9y=11,故答案为:11

已知x-y/x+y=3,求代数式5(x-y)/x+y-x+y/2(x-y)

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

已知:x-y=1,x+y=-7,求代数式3(x+y)+20(x-y)-5的值

x-y=1,x+y=-73(x+y)+20(x-y)-5=3×(-7)+20×1-5=-21+20-5=-6

已知5x=7y,x与y(  )

因为5x=7y,所以x:y=7:5=75(一定),所以x和y成正比例;故选:B.

已知:4x = 5y,求(x+y) :y

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

已知方程组{2x+y=7,x+2y=8,则x-y=(),x+y=().

解二元一次方程组,①*2,然后①-2得出X=2Y=3则X-Y=-1X+Y=5

(3x-4y)(5x-4y)-(7x-8y)(6y-5x)

(3x-4y)(5x-4y)-(7x-8y)(6y-5x)=15x²-12xy-20xy+16y²-(42xy-35x²-48y²+40xy)=15x²