已知(2x^2 2y^2)(x^2-2 y^)

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已知(2x^2 2y^2)(x^2-2 y^)
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-3y/y等于2 求x-y/2x 求x-3y/2x+y

(2x-3y)/y=2=>2x-3y=2y=>y=2/5x(1)将(1)式代入待求式子:(x-y)/2x=(x-2/5x)/2x=3/10同理可得(x-3y)/(2x+y)=-1/12

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*x-4xy+4y*y=0 求[2x(x+y)-y(x+y)]/(4x*x-4xy+y*y)的值?

即(x-2y)²=0x-2y=0所以x=2y所以原式=(2x²+2xy-xy-y²)/(4x²-4xy+y²)=(2x²+xy-y²

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,y满足约束条件:x-y+1>=0,x+y-2>=0,x

最小值0.5,1.5,-1最大值1,1,-1/3约束区域是一个三角形,把三角形的三个顶点代入.可以检验出最大值最小值.

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x²+y²+5=2x+4y,求【2x²-(x-y)(x-y)】【(x+y-1)(x-y

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=

已知集合A ={y \y =2~x ,x

呵呵..诚信的好孩子!我代表全宇宙人民向你致敬!(1)当a>0时0

已知x=1/3,y=-1/2,求代数式x-(x+y)+(x+2y)-(x+3y)+(x+4y)-(x+5y)+...-(

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