已知x^2 y^2 2x 8y-8=0 x^2 y^2-4x-4y-2=0

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已知y=根号(x-8)+根号(8-x)+18,求代数式[(x+y)/(根号x+根号y)]-2xy/(x根号y-y根号x)

y=根号(x-8)+根号(8-x)+18,x-8≥0,8-x≥0x=8,y=18[(x+y)/(根号x+根号y)]-2xy/(x根号y-y根号x)=26/(2√2+3√2)-288/(8*3√2-18

已知x^2+y^2-6x-8y+25=0,求分式y/x - x/y的值

x^2+y^2-6x-8y+25=0x^2-6x+9+y^2-8y+16=0(x-3)^2+(y-4)^2=0∴x=3且y=4y/x-x/y=4/3-3/4=7/12

已知x^2+y^2-6x-8y+25=0 求y/x-x/y的值

x^2+y^2-6x-8y+25=0[x-3]^2+[y-4]^2=0x-3=0===>x=3y-4=0===>y=4y/x-x/y=4/3-3/4=7/12

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+y/x=11/8,求x-y/x+2y的值

(x+y)/x=11/88x+8y=11y8x=3yy=8x/3(x-y)/(x+2y)=(x-8x/3)/(x+16x/3)=(-5x/3)/(19x/3)=-5/19

已知:3x=8y.求(1)x+y/y (2)2x+3y/x-2y

3x=8yx/y=8/3(1)x+y/y=x/y+1=8/3+1=11/3(2)2x+3y/x-2y分子分母同时除以y得=(2x/y+3)/(x/y-2)=(16/3+3)/(8/3-2)=(25/3

已知(x+y)(x+y+2)-8=0,则x+y的值是?

我教个更简单的方法令x+y=t所以原式=t(t+2)-8=0t²+2t-8=0(t-2)(t+4)=0t=2或t=-4所以x+y的值是2或-4

已知X-Y/X+Y=3,求代数式2(x-y)/X+Y-3X+Y/X+Y

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

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2x+2y=-8,4x+y=15求(x-y)^-(3x+2y)^

2x+2y=-8,.14x+y=15.21式*2-2式得3y=-21y=-7(x-y)^2-(3x+2y)^2=[(x-y)-(3x+2y)][(x-y)+(3x+2y)]=(-2x-3y)(4x+y

已知x.x+y.y+8x+6y+25=0,求代数式(x.x-4y.y)/(x.x+4xy+4y.y)-x/(x+2y)的

x平方+8x+16+y平方+6x+9=0(x+4)平方+(y+3)平方=0∴x+4=0y+3=0∴x=-4,y=-3原式=(x+2y)(x-2y)/(x+2y)平方-x/(x+2y)=(x-2y)/(

已知x,y满足x^2+y^2-6x+8y+25=0,求y^x的值

x^2+y^2-6x+8y+25=0x²-6x+9+y²+8y+16=0(x-3)²+(y+4)²=0x-3=0x=3y+4=0y=-4y^x=-4³

已知x+y/y=11/8,求x-y/x+2y的值.

x+y/y=11/8说明11y=8(x+y)即8x=3y所以x-y/x+2y=3(x-y)/3(x+2y)=(3x-3y)/(3x+6y)=(3x-8x)/(3x+16x)=-5/19

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

已知x^2+4y^2-4x+8y+8=0,求x+y

(x²-4x+4)+(4y²+8y+4)=0(x-2)²+4(y+1)²=0则:x-2=0且y+1=0得:x=2、y=-1x+y=1