化简x²-4x+4除2x

f(x)=a(2x^2-3x-2)+2x+3=a(2x=1)(x-2)+2x+3g(x)=a(2x^2-3x-2)+4x-1=a(2x=1)(x-2)+4x-1a能整除所以只要算(2x+3)/(2x+

f(x)=m(x)(x^2+2x+3)+(x+12)f(x)=n(x)(x+1)^2+(5x+4)=n(x)(x^2+2x+1)+(5x+4)=n(x)(x^2+2x+3)+x+12+4x-2n(x)

[(x-4)/(x^2-1)]÷[(x^2-3x-4)/(x^2+2x+1)]+1/(x-1)=[(x-4)/(x+1)(x-1)]÷[(x-4)(x+1)/(x+1)^2]+1/(x-1)=[(x-

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

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

有括号吗?如果没有括号（2x²-20x/x²-14x+4）就应该是（2x²-20/x-14x+4）（1-x-12/2-x）就应该是（1-x-6-x）=（-5-2x）

原式=(x-1)²/2(x²-9)÷[-3(x-1)/4x(x-3)]=(x-1)²/2(x+3)(x-3)×[-4x(x-3)/3(x-1)]=-2x(x-1)/3(x

设f(x)=(2x^2-3x-2)*R(x)+2x+3g(x)=(2x^2-3x-2)*Q(x)+4x-1（其中R(x),Q(x)均为正系数多项式）f(x)-g(x)=(2x^2-3x-2)*(R(x

由（x^4-x²-4x-4)/(x-2)(x+1)=[x²(x²-1)-4(x+1)]/(x-2)(x+1)=[x²(x-1)(x+1)-4(x+1)]/(x-

[x+1-15/(x-1)]÷[(x^2-4x)/(x^2-x)]=[(x^2-1-15)/(x-1)]÷[x(x-4)/x(x-1)]=[(x^2-16)/(x-1)]÷[(x-4)/(x-1)]=

原式=[(x+2)/x(x-2)-(x-1)/(x-2)²]*x/(x²-4)=[(x+2)(x-2)-x(x-1)]/x(x-2)²*x/(x+2)(x-2)=(x&#

原式=2(x-3)/(x-2)²*1/(x+3)*[-(x+3)(x-2)/(x-3)]=-2(x-3)(x+3)(x-2)/[(x-2)²(x+3)(x-3)]=-2/(x-2)

1.（4x^2-25y^2）/(2x-5y)=(2x+5y)(2x-5y)/(2x-5y)=2x+5y2.(2x^2-2x+1\2)/(x-1\2)=2(x^2-x+1/4)/(x-1/2)=2(x-

1.设多项式f(x)被x^2-1除后的余式为3x+4,并且已知f(x)有因式x,若f(x)被x(x^2-1)除后余式为px^2+qx+r,则p^2-q^2+r^2=设f(x)=(x^2-1)p(x)+

x÷x²+3x+2=2x+2-x÷x²+2x+1=x+1-1÷3x+6=?

7.2/4*x=4.321.8x=4.32x=2.41.8/5x=1/21.8x=2.5x=25/18

x减7.2除4等于x加9.6除8（X-7.2）/4=（X+9.6）/82X-7.2*2=X+9.6X=9.6+7.2*2X=24

x除35+2x-4除35-x除28+2x-1除35=12除60x/35+2x-4/35-x/28+2x-1/35=12/60x/35-x/28+4x-1/7=1/5等式两边都乘以7得x/5-x/4+2

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