和杨幂撞脸的女孩:设x=√3 -2,则x^6+3x^5+11x^3+2x+1=_____
来源:百度文库 编辑:高校问答 时间:2024/10/04 14:53:18
设x=√3 -2,则x^6+3x^5+11x^3+2x+1=_____
同上,需要步骤,谢谢
同上,需要步骤,谢谢
解:
x^6+3x^5+11x^3+2x+1
=x^3*(x^3+3x^2+11)+2x+1
=x^3*[(x^2*(x+3)+11]+2x+1
=x^3*[(x^2*(√3 -2+3)+11]+2x+1
=x^3*[(x^2*(√3 +1)+11]+2x+1
=x^3*[(√3 -2)*(√3 -2)*(√3 +1)+11]+2x+1
=x^3*[(√3 -2)*(1-√3)+11]+2x+1
=x^3*[(√3 -2)*(1-√3)+11]+2x+1
=x^3*[-5+3√3+11]+2x+1
=x^3*(6+3√3)+2x+1
=3x^3*(2+√3)+2x+1
=3*(√3 -2)*(√3 -2)*(√3 -2)*(√3+2)+2x+1
=-3*(√3 -2)*(√3 -2)+2*(√3 -2)+1
=-3*(7-4√3)+2√3 -3
=-24+14√3
你们学的那部分的知识?学没学因式整除?
设x=√3 -2,则x^6+3x^5+11x^3+2x+1=_____
x^6+x^5+x^4+x^3+x^2+x+1=0
1+x+x^2+x^3=0 ,求x+x^2+x^3+...+x^2000
x>0,f(x)=(x+1/x)^6-(x^6+1/x^6)-2(分子),(x+1/x)^3+(x^3+1/x^3)分母 则f(x)>=?
x>0,f(x)=(x+1/x)^6-(x^6+1/x^6)-2(分子),(x+1/x)^3+(x^3+1/x^3)分母 则f(x)?
解方程(x+1)(x+2)(x+3)(x+4)=(x+1)(x+1)+(x+2)(x+2)+(x+3)(x+3)+(x+4)(x+4)
(x+1)(x+2)(x+3)(x+4)=(x+1)(x+1)+(x+2)(x+2)+(x+3)(x+3)+(x+4)(x+4)
谁会解这道题:(X-5)/(X-7)+(X-2)/(X-4)=(X-3)/(X-5)+(X-4)/(X-6)求X的值
x*(x/3)=(x-5)*(x/3+1)/2 x=?
(x-4)(x-2)-(x-1)(x+3),其中x=-2/5