{█(1-x^2=y@1-y^2=z@1-z^2=x)┤

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{█(1-x^2=y@1-y^2=z@1-z^2=x)┤
[(2x+y)^2+(y+2x)(y-2x)-2y(4x-y)]/4y,其中x=1/2,y=1/3

[(2x+y)^2+(y+2x)(y-2x)-2y(4x-y)]/4y=(4x^2+4xy+y^2+y^2-4x^2-8xy+2y^2)/4y=(-4xy+4y^2)/4y=-x+y=-1/2+1/3

化简[(3x+4y)^2-(2x+y)(2x-y)+(-x+y)(5x-y)]除以-2y,其中x=-1,y=1

原式=(9x²+24xy+16y²-4x²+y²-5x²+6xy-y²)÷(-2y)=(30xy+16y²)÷(-2y)=-15x

2(x+y) 3x+3y=24 x+y/2x x y/2y= 1

由2(X+Y)3X+3Y=24得:2(X+Y)X+Y=8①;(X+Y/2X)XY/2Y=1得:X+Y=4②;由①、②得出Y=8(1-X),进入②知X=4/7;即Y=24/7

设x>1,y>0,若x^y+x^-y=2根号2,则x^y-x^-y等于

Dx^y+x^-y=2根号2===>(x^y+x^-y)^2=8===>x^2y+x^-2y+2=8===>x^2y+x^-2y=6(x^y-x^-y)^2=x^2y+x^-2y-2=6-2=4==>

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

1、x(x-y)(x+y)-x(x+y)^2

1)x(x-y)(x+y)-x(x+y)^2=x((x-y)(x+y)-(x+y)^2)=x(x^2-y^2-x^2-2xy-y^2)=x(-2xy-2y^2)=-2xy(x+y)2)(2a+b)(2

[(-x-y)(-x+y)-(x+y)^2-x(y-y^2)}÷1/2y

[(-x-y)(-x+y)-(x+y)^2-x(y-y^2)}÷1/2y=[x²-y²-x²-2xy-y²-xy+xy²]/(y/2)=[(x-2)y

(1)(x^2/x)-y-x-y

(1)x^2/x)-y-x-y=x-y-x-y=-2y(2)(a/a-b)-(a/a+b)-(2b^2/a^2-b^2)=a(a+b-a+b)/(a^2-b^2)-(2b^2/a^2-b^2)=2b/

先化简再求值(x-y)(x+y)-(x-2y) 的完全平方+x(3x-5y)-(x-y)(x-2y),其中x=1/2 y

解(x-y)(x+y)-(x-2y)²+x(3x-5y)-(x-y)(x-2y)=(x²-y²)-(x²-4xy+4y²)+(3x²-5xy

先化简,再求值[(2x+y)^2-(2x+y)(2x-y)-y(5x+y)]÷(1/2y),其中x-y=2

再答:再答:看后面这个图再答:不懂可以问我

{3(x+y)-4(x-y)=4 {x+y/2 + x-y/6=1

3(x+y)-4(x-y)=4(x+y)/2+(x-y)/6=1令a=x+y,b=x-y3a-4b=4(1)a/2+b/6=1则3a+b=6(2)(2)-(1)5b=2b=2/5a=(6-b)/3=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/

先化简,再求值:[(x+y)(x-y)-(x-y)^2+2y(x-3y)]/(-4y),其中x=1,y=-2

为你提供精确解答先化简:(x^2-y^2-x^2+2xy-y^2+2xy-6y^2)/(-4y)=(4xy-8y^2)/(-4y)=-x+2y=-1-4=-5其他的正在为你解答.

y''+y'=y'y x=2时,y=x,y'=1/2,求通解

令y'=p,那么y"=dp/dx=dp/dy*dy/dx=p*dp/dy所以原方程可以化为p*dp/dy+p=py即dp=(y-1)*dy等式两边积分得到p=y'=0.5y^2-y+C(C为常数)x=

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=

若|x+2y-1|+y²+4y+4=0,求(2x-y)²-2(2x-y)(x+2y)+(x+2y)&

∵|x+2y-1|+y²+4y+4=0∴|x+2y-1|+(y+2)²=0∴x=5,y=-2(2x-y)²-2(2x-y)(x+2y)+(x+2y)²=[(2x

[(y-2x)(-2x-y)-4(x-2y)²]*2y,其中x=1,y=2

x)(-2x-y)-4(x-2y)²]*2y,其

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