(2x)÷(1+x的平方)的极值点和极值-搜答案-拍题作业网

x的平方+x分之1+x的平方+3x+2分之1+x的平方-2x分之2=1/(x²+x)+1/(x²+3x+2)+2/(x²-2x)=1/x(x+1)+1/(x+1)(x+2

x(x²-2x+1)-3x(x²-4x+1)=x³-2x²+x-3x³+12x²-3x=-2x³+10x²-2x=-2x

x的平方+x+2

1.无解2.任意实数

是不是[3/(x-1)-x-1]÷[(x-2)/(x²-2x+1)][3/(x-1)-x-1]÷[(x-2)/(x²-2x+1)]={[3-(x²-1)]/(x-1)}÷

原式=-2x³+6x²-3x³-3x²=-5x³+3x²

=-(x-1)/(x+2)(x-1)×(x+2)/(x+1)(x-1)=-1/(x+1)(x-1)=-1/(x²-1)=1/(1-x²)

=[(x+2)/x(x-2)-(x-1)/(x-2)²]×(x-4)/x=[(x-2)(x+2)-x(x-1)]/x(x-2)²×(x-4)/x=(x-4)/x(x-2)²

1：2分之3X的平方-{-2分之1X平方}+{-2X平方}=3x²/2+x²/2-2x²=4x²/2-2x²=2x²-2x²=02

x的平方-6x+9/9-x的平方÷2x-6/x的平方+3x{(x-3)²/[(3-x)(3+x)]}÷{2(x-3)/[x(x+3)]}=[(3-x)/(3+x)]÷{2(x-3)/[x(x

x的平方-2x+1分之x-3÷x-1分之x的平方-9=[(x-3)/(x-1)²]*[(x-1)/(x+3)*(x-3)]=1/(x-1)(x+3)

原式=(x+1)²(x-1)²-4(x+1)²=(x+1)²[(x-1)²-4]=(x+1)²(x-1+2)(x-1-2)=(x+1)&su

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

-（x+2)的平方+16（x-1)的平方=16(x-1)^2-(x+2)^2=(4x-4)^2-(x+2)^2=(4x-4+x+2)(4x-4-x-2)=(5x-2)(3x-6)

(9x平方-6x+1)÷(x-2)×(2x-(x)的平方)÷(3x-1)=(3x-1)²÷(x-2)×(2-x)x÷(3x-1)=(3x-1)²÷(3x-1)×(2-x)x÷(x-

原式=[(x+2)/x(x+2)-(x-1)/(x-2)²]÷[(x²-4)/x]×(x-2)²=[x-(x-1)/(x-2)²]×[x/(x+2)(x-2)]

4x²-3x

x^2-2=0x^2=2原式=x^2-2x+1+x^2/x+2=2-2x+1+2/x+2=(2x+4-2x^2-4x+x+2+2)/x+2=(2x+4-4-4x+x+2+2)/(x+2)=4-x/x+

x^2+3x+1=0显然x!=0x+3+1/x=0x+1/x=-3(x+1/x)^2=9x^2+1/x^2+2=9x^2+1/x^2=7x^2+1/x^2+2x+2/x=7+2*(-3)=1

9+6+1/9-3+1=13又9分之1

原式=[2x(x+1)/(x+1)(x-1)-x(x-1)/(x-1)²]÷x/(x+1)=[2x/(x-1)-x/(x-1)]×(x+1)/x=x/(x-1)×(x+1)/x=(x+1)/