若x>0,则x-1 6x^3-搜答案-百度作业网

x²+3x-1=0x²+3x=1(x-3)/(3x^2-6x)除以[x+2-5/(x-2)]=(x-3)/3x(x-2)÷(x²-4-5)/(x-2)=[(x-3)/3x

f（x）＝｛2x,x＜0x²,x≧0｝因为f(x)=161.2x=16x=8>0,矛盾2.x²=16x=4或-4（舍去）所以x=4

x*x+x+1=0(x-1)(x*x+x+1)=0x^3-1=0x^3=1x^7+2x^6+x^5+x^4+x^3+x*x+1=x^6*x+2x^6+x^3*x^2+x^3*x+x^3+x*x+1=x

(2007+1)÷4=502∴1+x+x^2+x^3+x^4+x^5+...+x^2007+x^2008-x^2008=1+x+x²+x³+x^4(1+x+x²+x

提公因式法吧x^97+x^98+x^99+x^100+x^101+x^102+x^103=x^97(1+x+x^2+x^3+x^4+x^5+x^6)因为1+x+x^2+x^3＝0所以原式＝x^97(x

x²+3x=1所以原式=x³+3x²+2x²+6x-x+8=x(x²+3x)+2(x²+3x)-x+8=x+2-x+8=10

x²+3x+1=0则x²+1=-3x两边同时除以x,得x+1/x=-3

x³+x²+x+1=0四个一组原式=x^2009(x³+x²+x+1)+x^2005(x³+x²+x+1)+……+x(x³+x&#

（x-3)/（3x^2-6x）/[x+2-(5)/(x-2)]=（x-3)/(3x)/（x-2）/[x+2-(5)/(x-2)]=（x-3)/(3x)/[(x+2)(x-2)-5]=（x-3)/(3x

题目不清晰啊是x-1/（x²+2x+1）-(x+1)/(x-1)还是x-1/(x²)+2x+1-x+1/(x-1）还是其他什么的?分子分母是什么请用括号括起来~再问：(x-1)/(

3x^2+6x-1=3x^2+6x+9-9-1=(3x^2+6x+9)-9-1=3(x^2+2x-3)-10=-10

X^2+3x-1=0可得X^2+3x=1x^3+5x^2+5x+18=x^3+3x^2+2x^2+5x+18=x(x^2+3x)+2x^2+5x+18=x+2x^2+5x+18=2x^2+6x+18=

解代数式,代入=1

=x^2005*(x^3+x^2+x+1)=0再问：不太明白，过程能不能在具体些再答：=x^2005*x^3+x^2005*x^2+x^2005*x+x^2005*==x^2005*(x^3+x^2+

1,1/32,m=a^2+2b^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

若x²+3x-1/3=0则代数式x-3/3x²-6x÷(x+2-5/x-2)的值(x-3)/(3x²-6x)÷[(x+2)-5/(x-2)]=(x-3)/3x(x-2)÷

函数为奇函数根据a+b>=2ab得出当x>0时f(x)>=72又因为函数为奇函数所以f(-x)=-f(x)所以当x

∵x^2-3x+1=0∴x^2+1=3X∴（2x^5-5x^4+2x^3-8x^2)/(x^2+1)=[2X^3(X^2-3x+1)+x^2(x^2-3x+1)+3x(x^2-3x+1)-3x]/3x

x^2+x=1x^2=1-x2x^3-3x^2-x=2x(x^2+x)-5x^2-x=2x-5(1-x)-x=6x-5.