x*=y-3

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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=

6(x-y)-2(x+y)=14 3(x-y)+(x+y)=5

设x-y=mx+y=n则6m-2n=143m+n=5解得m=2n=-1即x-y=2x+y=-1解得x=1/2y=-3/2

x+y/2+x-y/3=6,4(x+y)-3(x-y)=-20

由(1)得3x+3y+2x-2y=365x+y=36(3)由(2)得4x+4y-3x+3y=-20x+7y=-20(4)(3)×7-(4)得34x=272∴x=8把x=8代入(3)得y=-4∴x=8y

{(x+y)/2+(x-y)/3=6 4(x+y)-3(x-y)=-20

{(x+y)/2+(x-y)/3=63(x+y)+2(x-y)=36(1)4(x+y)-3(x-y)=-20(2)由(1)*3+(2)*2得9(x+y)+6(x-y)+8(x+y)-6(x-y)=36

已知x/5=y/3,则(x/x+y)+(y/x-y)-(yxy/xxx-yxy)

y=3x/5原式=x/(x+3x/5)+(3x/5)/[x-3x/5]-(9x^3/25)/(x^3-9x^3/25)=8/3-3/2-9/16=29/48

3(x+y)-4(x-y)=11,2(x-y)+5(x+y)=27

化简得:-x+7y=11①7x+3y=27②①式×7得:-7x+49y=77③②+③得:52y=104∴y=2代入①得:x=3∴x=3,y=2再问:亲,是代入法哦!再答:代入法①式得3x+3y-4x+

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

4(x+y)-3(x-y)=-20,2/x+y+3/x-y=6

第二个方程是不是写错了2/(x+y)+3/(x-y)=6是这样吗

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

{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

若(x-y)/(x+y)=3 则(x+2y)/(6x-7y)等于

若(x-y)/(x+y)=3那么x-y=3(x+y)x-y=3x+3y故x=-2y所以(x+2y)/(6x-7y)=(-2y+2y)/(6*(-2y)-7y)=0如果不懂,请Hi我,祝学习愉快!

(5x+3y)(3y-5x)-(4x-y)(4y+x)=

(5x+3y)(3y-5x)-(4x-y)(4y+x)=(3y)^2-(5x)^2-(4x^2+15xy-4y^2)=9y^2-25x^2-4x^2-15xy+4y^2=13y^2-15xy-29x^

{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

已知|3x-y|+|x+y|=0,求x-y除以x*y的值.

绝对值项恒非负,两绝对值项之和=0,两绝对值项分别=03x-y=0(1)x+y=0(2)(1)+(2)4x=0x=0,代入(2)y=-x=0(x-y)/(xy)无意义,因此题目错了.

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

(x+y)(x+2y)(x+3y)(x+4y)=-40

我把方法告诉你,最后的答案你自己做吧,很容易.(x+y)(x+2y)(x+3y)(x+4y)=-40(x+y)(x+4y)(x+2y)(x+3y)=-40(x^2+5yx+44)(x^2+5yx+66