3(x-1)=y+5 5(y-1)=3(x 5)用代入法解下列方程组

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3(x-1)=y+5 5(y-1)=3(x 5)用代入法解下列方程组
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

{3(x+y)+3(y+x)=1,3(x+y)+4(y-x)=-1

3(x+y)+3(y-x)=1(1)3(x+y)+4(y-x)=-1(2)(2)-(1)得y-x=-2(3)代入(1)3(x+y)-6=1x+y=7/3=>x=7/3-y又由(3)得x=y+2y+2=

x>1,y>0,且满足xy=x^y,x/y=x^3y,求x+y

x=4,y=0.5,x+y=4.5(与人家的做法一样……)(1)解题思路是以S3为基准,用S3表示出S1,S2,S4即可.在三角形BCD中有:S2/S3=DF/CF,故S2=(DF/CF)S3;同理,

计算:(x-y)(1/9x-y)-(1/3x-y)(1/3x+y)

1/9x^2-10/9xy+y^2-(1/9x^2-y^2)=-10/9xy+2y^2=2y(y-5/9x)

设变量x,y满足约束条件x-y>-1,x+y>1,3x-y

是11过程就不说了画出坐标图就可以了、

(x+2)(x+3) (x-4)(x+1) (y+4)(y-2) (y-5)(y-3)

(x+2)(x+3)=x²+5x+6(x-4)(x+1)=x²-3x-4(y+4)(y-2)=y²+2y-8(y-5)(y-3)=y²-8y+15x²

已知x-y=1,y≠0,求{(x+2y)²+(2x+y)(x+4y)-3(x+y)(x-y)}÷y的值.

已知x-y=1,则y=x-1,x=y+1{(x+2y)²+(2x+y)(x+4y)-3(x+y)(x-y)}÷y=(x²+4xy+4y²+2x²+9xy+4y&

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

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

计算:1/(x-y)-1/(x+y)-2y/(x^2+y^2)-4y^3/(x^4+y^4)-8x^7/(x^8+y^8

1/(x-y)-1/(x+y)-2y/(x^2+y^2)-4y^3/(x^4+y^4)-8y^7/(x^8+y^8)=【(x+y)-(x-y)】/(x-y)(x+y)-2y/(x^2+y^2)-4y^

{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

已知实数x,y满足x−x+1=y+3−y

∵x−x+1=y+3−y,∴x+y=x+1+y+3≤2x+y+42则(x+y)2≤2(x+y+4)解得:-2≤x+y≤4∴x+y的最大值为4故答案为:4

1、(-7x^y)(2x^y-3xy^3+xy) 2、((x-y)^6)/((y-x)^3)/(x-y)

1、(-7x^y)(2x^y-3xy^3+xy)=-14x^(2y)+21x^(y+1)y^3-7x^(y+1)y;2、((x-y)^6)/((y-x)^3)/(x-y)=-(x-y)^3/(x-y)

{4/(x+y)+6/(x-y)=3 {9/(x-y)-1/(x+y)=1

完整设1/(x+y)=a,1/(x-y)=b原方程组可变为4a+6b=39b-a=1a=9b-136b-4+6b=3b=1/6,a=1/2x+y=2x-y=6所以原方程组的解为:x=4,y=-2

y'=(x-y+1)/(x+y-3)通解

(x+y^2+3)dy=(x-y+1)dx或:xdy+ydx+(y^2+3)dy-(x+1)dx=d(xy)+(y^2+3)dy-(x+1)dx=0通解为:xy+y^3/3+3y-x^2/2-x=C