y=(x∧2-1) x∧2-3x 2

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y=(x∧2-1) x∧2-3x 2
若变量x,y满足约束条件{x>=-1;y>=x;3x+2y

用图像法,画x>=-1;y>=x;3x+2y

分解因式计算:x^2(x-y)+(y-x)=?x^3+x^2+1\4x=?(2x+y)^2-(x-2y)^2=?

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

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

y=(x-1)(x-2)(x-3)(x-4)(x-5)(x-6)(x-7)(x-8)(x-9)(x-10)的导数在x=1

设a=(x-1)(x-2)(x-3)(x-4)(x-5)(x-6)(x-7)(x-8)(x-9)那么y=a*(x-10);那么y^=a^*(x-10)+a*(x-10)^=a^*(x-10)+a那么y

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

1.分离法y=(3x+1)/(x-2),定义域为{x|x≠2}y=(3x+1)/(x-2)=7/(x-2)+3当x≠2时,7/(x-2)≠0,y≠3∴函数值域为{y|y∈R且y≠3};2.利用函数的单

y=(x^2+x)/(x+1)

这样算,分离变量:x²+x=(x+1)²-(x+1)然后,除下来,就等于x+1-1=x注意,x≠-1!

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

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

函数,y=3x/(x^2+x+1) ,x

y=3/(x+1/x+1)x+1/x≤-2,所以x+1/x+1≤-1令t=x+1/x+1,则t≤-1,y=3/t值域为[-3,0)再问:你写的我看不大懂再问:一步步写再答:

(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(x+2y)∧2-(x+y)(3x-y)-5y∧2,其中x+-2,y=1/2

原式=-2x(x-2y)把x=-2,y=1/2代入得原式=-12再问:我邻居说是等于-14,你有没有写错啊再答:没错再问:我看看过程再答:(x+2y)²-(x+y)(3x-y)-5y&sup

{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2x-3y+4=0则x(x*x-1)+x(5-x*x)-6y+7

x(x*x-1)+x(5-x*x)-6y+7=-x+5x-6y+7=2(2x-3y)+7=2*(-4)+7=-8+7=-1再问:能在写详一点吗-x+5x-6y+7再答:x(x*x-1)+x(5-x*x

函数y=3x/(x^2+x+1) (x

原式可以化为:y*x^2+(y-3)*x+1=0Δ=(y-3)^2-4y≥0解得y≥9或y≤1由于x

9y=3x-(x-1),4(x+y)-(2x+4)=8y

解9y=3x-(x-1)2x-9y+1=0①4(x+y)-(2x+4)=8y2x-4y-4=0②②-①得:5y-5=0∴y=1将y=1代入①得:2x-9+1=0∴x=4∴方程的解为:x=4,y=1

y=2x∧2+2x+3/x∧2+x+1求值域,

y=2x∧2+2x+3/x∧2+x+1=2x∧2+2x+2+1/x∧2+x+1=2+1/x∧2+x+1∵x^2+x+1=x^2+x+1/4+3/4=(x+1/2)^2+3/4>=3/4∴0再问:倒数第

y=(2x*x-2x+3)/(x*x-x+1),求值域?

y=2x2;-4x+2-3=2(x-1)2;-30≤x≤3-1≤x-1≤2所以0≤(x-1)2;≤40≤2(x-1)2;≤8-3≤2(x-1)2;-3≤5所以值域[-3,5]

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