y=(x^2-1) (x^2-3x 2),x=1是间断点-搜答案-搜狗做题网

用图像法,画x>=-1;y>=x;3x+2y

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

(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+

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=(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=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

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

（2x+(y-1)）(2x-(y-1))=4x^2-(y-1)^2(3x+y)(3x-y)=9x^2-y^2相减得4x^2-y^2+2y-1-9x^2+y^2=-5x^2+2y-1带入得-5*1/25

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：

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

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

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

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=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]

(x-2y)/(x+2y)=3取倒数(x+2y)/(x-2y)=1/3所以原式=(1/3)[(x-2y)/(x+2y)]-3[(x+2y)/(x-2y)]=(1/3)×3-3×(1/3)=0

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

y=(x^2-1) (x^2-3x 2),x=1是 间断点