y=(x^2-4)/(x^2-5x+6) 在x=2,x=3处间断,说明间断点的类型,如果是可去间断点,补充或改变函数的定义使它连续
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y=(x^2-4)/(x^2-5x+6) 在x=2,x=3处间断,说明间断点的类型,如果是可去间断点,补充或改变函数的定义使它连续
y=(x^2-4)/(x^2-5x+6) 在x=2,x=3处间断,说明间断点的类型,如果是可去间断点,补充或改变函数的定义使它连续
y=(x^2-4)/(x^2-5x+6) 在x=2,x=3处间断,说明间断点的类型,如果是可去间断点,补充或改变函数的定义使它连续
因在x=2这个点处用络必达法则计算知为可去间断点(左右极限相等),y(2)=-4
x=3时为跳跃间断点(左右极限不相等),y(3)=0
怎么不给分啊,同学
若2x-3y+4=0则x(x*x-1)+x(5-x*x)-6y+7
xy²/(x²y-y) × x²/(x²+x)=(x²-3x)/(x²-5x) × 2x-10/(x²-6x+9)=化简求植x²-1/(x²-x-2)除以x/2x-4,其中x=1/2[x-x/(x+1)]除以[x/(2x-4)] ,其中x=√2+1
{x+y=1 ,xy=-6{x(2x-3)=0,y=x²-1{(3x+4y-3)(3x+4y+3)=0,3x+2y=5{(x-y+2)(x+y)=0,x²+y²=8{(x+y)((x+y-1)=0,(x-y)(x-y-1)=0
6-4x-x^2=5/|sin(y/x)cos(y/x)|的解
x+y/2+x-y/6=5 2(x+y)-3(x-y)=-2
把(x-y)看作一个整体,化解:5(x-y)+2(x-y)-4(x-y)= 3(x-y)把(x-y)看作一个整体,化解:5(x-y)+2(x-y)-4(x-y)=3(x-y)²-4(x-y)+7(x-y)-6(x-y)²=
(x+y)/2- x-y/2=6 3(x+y)=4(x-y)
x+y/3+x-y/2=6 4(x+y)-2(x-y)=40
{3(x+y)-4(x-y)=4 {x+y/2 + x-y/6=1
3(x+y)+2(x-y)=36 4(x+y)-5(x-y)=2
{6(x-y)-7(x+y)=21 {2(x-y)-5(x+y)=-1
6(x-y)-2(x+y)=14 3(x-y)+(x+y)=5
x+y/2+x-y/3=6,4(x+y)-3(x-y)=-20
{(x+y)/2+(x-y)/3=6 4(x+y)-3(x-y)=-20
4(x+y)-3(x-y)=-20,2/x+y+3/x-y=6
x*x+2x+y*y-4y+5=0 x= y=
4(x-y)-3(x-y)=9 5(x-y)+2(x-y)=11
3(x+y)-4(x-y)=11,2(x-y)+5(x+y)=27