若x^2 3x-1=0,则x^3 5x^2 5x 8
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答:x+1/x=3x^2/(x^4+x^2+1)=1/(x^2+1/x^2+1)=1/[(x+1/x)^2-1]=1/(3^2-1)=1/8
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
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
每四项为一组,第一组x+x*x+x*x*x+x*x*x*x=x(x+x*x+x*x*x)=0以此类推结果为〇
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
x²-5x-1=0x²-1=5xx-1/x=5x²+x+1/x²-1/x=(x²+1/x²)+(x-1/x)=(x-1/x)²+2
(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^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)÷
/>x³+x²+x+1=0x²(x+1)+(x+1)=0(x+1)(x²+1)=0∵x²+1>0∴x+1=0,那么x=-1∴原式=-1+1-1+1-1
∵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.