B={(x,y) 2 x-3 y-4 =

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B={(x,y) 2 x-3 y-4 =
3a(x+y)-2b(x+y)=(x+y)(?)

可以提公因式x+y就得到了(x+y)(3a-2b)再问:-3的n次方b的n次方+1-6a的n次方=-3a的n次方b的n次方(?)再答:你问的这个我看不懂啊,根据你打的字,我完全写不出来你想表达的意思再

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

A=2x²-3xy+y²-x+2y,B=4x²-6xy+2y²+3x-y,若x=

是不是-2A+B=2a所以-4x²+6xy-2y²+2x-4y+4x²-6xy+2y²+3x-y=5x-5y=5a+15=2aa=-3A=18-27+9+3-6

(2x+y)(2x-3y)+x(2y+y) 2a(x-y)-4b(x-y)

(2x+y)(2x-3y)+x(2y+y)=(2y+y)(2x-3y+x)=(2y+y)(3x-3y)=3(2y+y)(x-y)2a(x-y)-4b(x-y)=2(x-y)(a-2b)

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+

x²(x+y)(y-x)-xy(x+y)(x-y) 已知a+b=4,ab=-2,求多项式3a²b+3

再问:-ab(a-b)²+a(b-a)²-ac(a-b)²再答:

若向量x,y满足3x+4y=a,2x+3y=b,a,b为已知向量,则x=?,y=?

联立方程3x+4y=a,2x+y=b,我们按解二元一次方程组的方法解的y=3b-2a,a=3a-4b.希望能给你带来帮助

如果分式3x-6y/x+y=0,那么x,y 应满足 A、x=2y B、x≠-y C、x=2y且x≠-y D、x=2y且y

c∵x+y≠0∴x≠-y∵3x-6y/x+y=0∴3x-6y=0∴x=2y∴x=2y且x≠-y

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

设A={(x,y)|y=-2x+3,x,y∈R},B={(x,y)|y=(4-3m^2)x+1,x,y∈R},求A∩B≠

不难看出,方程y=-2x+3和y=(4-3m^2)x+1都是直线方程,A∩B≠空集也就是两直线没有交点,同一平面内没有交点就是平行,也就是斜率相等-2=4-3m^2,m^2=2,m=±√2

已知集合A={(x,y)| (y-3)/ (x-1)=2,x,y∈R},B={(x,y)| 4x+ay=16,x,y∈

∵(y-3)/(x-1)=2,x≠1,可化为y-2x=1,∵A∩B=空集,∴y-2x=1方程组无解ay+4x=16∴1/a=-2/4≠1/16解得a=-2又x≠1,则y≠3,∴3a+4≠16,a≠4当

设A=2x²-3xy+y²-x+2y,B=4x²-6xy+2y²-3x-y,若x

设A=2x²-3xy+y²-x+2y,B=4x²-6xy+2y²-3x-y,且B-2A=a所以a=B-2A=4x²-6xy+2y²-3x-y

{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

设A=2x²-3xy+y²-x+2y,B=4x²-6xy+2y²+3x-y.若x

∣x-2a∣+(y+3)²=0因为∣x-2a∣≥0,(y+3)²≥0所以∣x-2a∣=0,(y+3)²=0所以x=2a,y=-3B-2A=(4x²-6xy+2y

1.设A={(x,y)/4x+y=6} B={(x,y)/3x+2y=7} 则A∩B等于多少

求A∩B即是求两直线的焦点,将两方程联立4X+y=63x+2y=7解得x=1y=2所以综上A∩B={(1,2)}

已知集合A={Y|Y=X^2-4X+3},B={Y|Y=-X^2-2X+2}则A∩B=

y=(x-2)²-1(x-2)²≥0所以(x-2)²-1≥-1同样(x+1)²≥0-(x+1)²≤0-(x+1)²+3≤3

已知集合A={(x,y)|2x+3y=6},B={(x,y)|x-2y=-4},求A∩B,

解方程组:2x+3y=6x-2y=-4==>2x-4y=-8两式相减:7y=14,y=2x=2y-4=0所以A∩B={(0,2)},

(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