已知x-2y=z 2x 2y=1 2x-y=z

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已知2x+y=4,求代数式[(x+y)2-(x-y)2-2y(x-12

原式=[x2+y2+2xy-x2-y2+2xy-2xy+y2]÷4y=(2xy+y2)÷4y=14(2x+y)=14×4=1.

已知2/x+2/y=根号24,求 x/y(x-y) - y/x(x-y)的值

2/x+2/y=根号24(2y+2x)/xy=2√6x+y=xy√6x/y(x-y)-y/x(x-y)=1/(x-y)[(x/y-y/x)]=1/(x-y)[(x²-y²)/xy]

已知2/x 2/y=根号24,求 x/y(x-y) - y/x(x-y)的值

你的题目有点问题我这样做了x/{y(x-y)}-y/{x(x-y)}=(x平方-y平方)/{xy(x-y)}=(x+y)/xy2/x+2/y=2(x+y)/xy=根号24

已知2x-y=10,求[(x²+y²)-(x-y)²+2y(x-y)]/4y

先化简[(x²+y²)-(x-y)²+2y(x-y)]/4y[(x²+y²)-(x-y)²+2y(x-y)]/4y=[x²+y&s

已知2x-y/x+3y=2,求代数式4x-2y/x+3y减4x+12y除以2x-y

2x-y/x+3y=2得x+3y/2x-y=1/24x-2y/x+3y减4x+12y除以2x-y=2(2x-y)/x+3y-4(x+3y)/2x-y=2乘以2-4乖以1/2=4-2=2

已知x/y=2,求2x(x+y)-y(x+y)/4x²-4xy+y²

原式=(2x-y)(x+y)/(2x-y)^2=(x+y)/(2x-y)x/y=2x=2y原式=3y/3y=1

高数:已知f(x+y,y)=x^2+y^2,求f(x,y)

这道题实际就是要把x^2+y^2变换成只由x+y和y组成的多项式x^2+y^2=x^2-y^2+2y^2=(x+y)(x-y)+2y^2=(x+y)[(x+y)-2y]+2y^2将式中(x+y)替换为

数学题已知p=x^/x-y-y^/x-y,q=(x+y)^-2Y(X+Y),

你的题是什么意思?“^/”是什么意思?

已知2x-y/x+3y=2,求代数式4x-2y/x+3y减4x+12y/2x-y的值

答案是2.4x-2y/x+3y=(2x-y/x+3y)乘以2=4,4x+12y/2x-y=(x+3y/2x-y)乘以4=1/2x4=2,所以代数式等于4-2=2.

已知X.x+y.y-4x+2y+5=0,求4X.X-12X Y+9Y.y的值

∵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&#

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x*x+4x+y*y-2y+5=0,则x*x+y*y=?

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

已知x+y=2,求12x

∵x+y=2,∴12x2+xy+12y2=12(x2+2xy+y2)=12(x+y)2=12×22=2;

已知4x=9y求(1)x+y/y (2)y-x/2x

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/

已知x^2+y^2-10x-12y+61=0,求x/y-y/x的值

x^2+y^2-10x-12y+61=0(x²-10x+25)+(y²-12y+36)=0(x-5)²+(y-6)²=0因为(x-5)²与(y-6)&

已知x^2=12,y^2=3,且|x-y|=y-x,求x+y的值.

x^2=12x=±2√3y^2=3y=±√3|x-y|=y-x说明y-x≥0即y≥x所以x不可能等于2√3,即x=-2√3若y=-√3,x+y=-3√3若y=√3,x+y=-√3

已知集合A ={y \y =2~x ,x

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已知实数x,y满足|x|+x+y=10、x+|y|-y=12,则x+y的值?

若x≤0,|x|=-x|x|+x+y=10y=10代入x+|y|-y=12得x=12>0矛盾,∴x>02x+y=10①若y≥0,x+|y|-y=x=12y=10-2x∴yx-2y=12②联立①②解得x

已知x^2-y^2=12,x-y=2,则x除以y=

(x+y)*(x-y)=12除以第二个式子得x+y=6,就可以求出x=4,y=2,所以答案为2