x的四次方加x的二次方加1分之想x的四次方

有没有括号?是不是这样的：1+(2x^4)/(2x^3)+x^2=x^2+x+1

X的二次方加X加1的和分之X=m左边分式的分子分母同除以x1/(x+1+1/x)=mx+1/x=1/m-1两边平方,x²+1/x²+2=1/m²+1-2/mx²

X方-X+1分之X=7(x²-x+1)/x=1/7x-1+1/x=1/7x+1/x=8/7X四次方+X二次方+1分之X二次方分子分母同时除以x²=1/(x²+1+1/x&

x^2-13x+1=0x1=[13+√(13*13-4)]/2=6.5+0.5√165x2=6.5-0.5√165当x＝x1时x^2=6.5x6.5+165/4+6.5√165=(169+165)/4

看你的题目都看了很少时间(x-4)/(x^2-1)÷(x^2-3x-4)/(x^2+2x+1)+1/(x-1)=(x-4)/(x^2-1)[(x^2+2x+1)/(x^2-3x-4)]+1/(x-1)

X^2+X^-2=3则：（X^2+X^-2）^2=9得到X^4+2+X^-4=9解得X^4+X^-4=7注：X^-2为X的-2次方,X^2为X的平方.望采纳

(x-2)^4=ax^4+bx^3+cx^2+dx+e令x=1则a+b+c+d+e=a*1^4+b*1^3+c*1^2+d*1+e=(1-2)^4=1令x=-1则a-b+c-d+e=a*(-1)^4+

x^2/(x^4+x^2+1)=1/(x^2+1+1/x^2)x^2-4x+1=0x-4+1/x=0x+1/x=4原式=1/[(x+1/x)^2-2+1]=1/(16-2+1)=1/15

x的平方加x的负二次方的值为7,x的四次方加x的负四次方的值为47再问：有过程吗

先分解因式,后乘法.(X^4-Y^4)/(X^3+X^2Y+XY^2+Y^3)=(X-Y)(X+Y)(X^2+Y^2)/(X^3+X^2Y+XY^2+Y^3)=(X-Y)(X^3+X^2Y+XY^2+

x²-4x+1=0两边同时除以xx-4+1/x=0x+1/x=4两边同时平方x²+2+1/x²=16x²+1/x²=14两边同时平方x^4+2+1/x

x^4+4=(x^2)^2+2^2+4x^2-4x^2=(x^2+2)^2-4x^2=(x^2+2)^2-(2x)^2=(x^2+2x+2)(x^2+2-2x)

x^4+y^4=x^4+2x^2y^2+y^4-2x^2y^2=(x^2+y^2)^2-2x^2y^2=(x^2+y^2+√2*xy)(x^2+y^2-√2*xy)

（2X-1）^5=（aX）^5+（bX）^4+（CX）^3+（dX）^2+eX+f,解得f=（5+a+b+c）X^2+dX^2+eX

由x³+x²+x+1=0,可得x=0所以1+x+x²+x³+x^4=1+0=1或者变形：1+x+x²+x³+x^4=1+（1+x+x

原式=[x/(x-y)]*[y^2/(x+y)]-[x^4y/(x^2+y^2)(x^2-y^2)]/[x^2/(x^2+y^2)]=xy^2/(x^2-y^2)-x^2y/(x^2-y^2)=xy(

/>∵x/(x²-x+1)=1/6∴(x²-x+1)/x=6x²-x+1=6xx²+1=7x∵(x²+1)×1/x=7x×1/xx²×1/x

x^4+2x+1=x^4-x²+x²+2x+1=x²(x-1)(x+1)+(x+1)²=(x+1)(x³-x²+x+1)

（3x+1）^4=81x^4+108x^3+54x^2+12x+1(运用二项式展开式)自己对照起来其他的问题就好办了主要是你表述的不太清楚

x^4+4=(x^2)^2+2^2+4x^2-4x^2=(x^2+2)^2-4x^2=(x^2+2)^2-(2x)^2=(x^2+2x+2)(x^2+2-2x)