lim2x+32x+1

求极限lim2x^3-x+1=?趋向无穷你确认这个等式是正确的吗？那就是+∞+正无穷lim2x^3-x+1=2

当x趋向于无穷大,求lim2x^3-x+1,无穷大,

求极限lim2x(3x^3+x+2)/(2x^2+1)(3x^2+4)化简得（6x^4+2x^2+4x）/(6x^4+11x^2+4)当x趋近于0时,极限为0当x趋近于无穷时,式子每项除以想x^4得{6+2/x^2+4/x^3}/{6+11

求极限:x趋于无穷大,lim2x^2-3x-4/√x^4+1（2x²-3x-4)/√(x^4+1)=[2-3/x-4/x²]/√(1+1/x^4)x趋于无穷大=[2-0-0]/√(1+0)=2/1=2x趋于无穷大，2x^

当x趋向于无穷大,求lim2x^3-x+1,答案1/2,为什么可能么,还有分母呢吧?看清楚

求极限lim2x-1/x^2x趋于∞rt告诉我为何极限为零.谢谢^O^应该是（2x-1）/x^2吧原式=2/x-1/x^2每一部分取极限都是零（常数/无穷=0）所以加起来为零

lim(2x+△x)=2?(当x趋向x0),有没人能给出计算过程.lim(2x+△x)=lim2x+lim△x=2limx+lim△x=?lim(2x+△x)=lim2x+lim△x=2limx+lim△x==0+2x->x0x0=0△x=

1)3-2x/1-x-1=x+2/x-32)x/3+x-x/2-x=23)x-1/x+1-x/x-1=4/x*-14)1/(x-1)(x-2)=1/(x-4)(x-5)麻烦帮帮忙、解下1)3-2x/1-x-1=x+2/x-35-4x-2/x

2x(7-3x)+5x(2x+1)=x(4x-3)+3214x-6x²+10x²+5x=4x²-3x+3222x=32∴x=16/112x(7-3x)+5x(2x+1)=x(4x-3)+3214x-6x^2+1

x=1-x/2+x/4-x/8+x/16-x/32 

x/x-4-32/x的平方+16=1x+1/x+5X/x+1=6不破

多项式x^2-3x+2,x-18x+32,x^2+x-6的公因式是（）A.x-1B.x-16C.x+3D.x-21:(x-2)(x-1)2:(x-2)(x-16)3:(x-2)(x+3)故选Dx^2-3x+2分解为(x-2)(x-1)x-1

8x^2(x+3)-8x(x+1)^2=8x^2-32x=4

x/x+(1-x)/x-1/xx/[x+(1-x)/(x-1/x)]=x/[x+(1-x)/((x²-1)/x)]=x/[x+x(1-x)/(x+1)(x-1)]=x/[x-x(x-1)/(x+1)(x-1)]=x/[x-x/(x

x+(1/2x-4)＝32-(2x+5)打开括号,合并同类项得1.5X-4=27-2X移项,合并同类项得3.5X=31解得X=62/7

计算:(x+1)(x^2+1)(x^4+1)(x^8+1)(x^16+1)(x^32+1)(1)当x不等于1(x+1)(x^2+1)(x^4+1)(x^8+1)(x^16+1)(x^32+1)=[(x-1)(x+1)(x^2+1)(x^4+

计算:(x+1)(x^2+1)(x^4+1)(x^8+1)(x^16+1)(x^32+1)(x+1)(x^2+1)(x^4+1)(x^8+1)(x^16+1)(x^32+1)=(x-1)(x+1)(x^2+1)(x^4+1)(x^8+1)(

解方程X-1=1/2X+1/4X+1/8X+1/16X+1/32XX-1=1/2X+1/4X+1/8X+1/16X+1/32XX-1=(1/2+1/4+1/8+1/16+1/32)XX-1=(31/32)XX-(31/32)X=1(1/32

解方程X-1=1/2X+1/4X+1/8X+1/16X+1/32XX-1=1/2X+1/4X+1/8X+1/16X+1/32XX-1=(1/2+1/4+1/8+1/16+1/32)XX-1=(31/32)XX-(31/32)X=1(1/32

x=1+1/2x+1/4x+1/8x+1/16x+1/32xx=1+1/2x+1/4x+1/8x+1/16x+1/32xx=1+x-1/32x1/32x=1x=32