已知x2-3x=1 =0,求x3+1/x3

来源:学生作业帮助网 编辑:作业帮 时间:2024/11/29 23:12:41
已知x2-3x=1 =0,求x3+1/x3
xRN@&&R'ef4@#&fAх ⃸0Q|3ӂ]4sϙ{{]ȋID9K-&86U+c*q{msѯ5Mp89XdL䒐{}9M2E#P:t&P3YX]k6[K&UCTxx} ay\5 ߀U](on "ҡնDᙆ'ˏ|Jx~Od{|ɧg a3aR>p3'&:,C'cUzb T "跎ÔNEҒ')Ka&aG3\fP*/Ɋ,b'0 wv)QtsQi wO\l}KS}Ek

已知x2-3x=1 =0,求x3+1/x3
已知x2-3x=1 =0,求x3+1/x3

已知x2-3x=1 =0,求x3+1/x3
x2-3x+1 =0
除以x得 x-3+1/x=0
x+1/x=3
x³+1/x³=(x+1/x)(x²-1+1/x²)
=3×[(x²+2+1/x²)-1-2]
=3×[(x+1/x)²-3]
=3×﹙3²-3)
=18

你好,很高兴回答你的问题

∵x²-3x+1=0
∴x+1/x-3=0
∴x+1/x=3
∴x²+1/x²=9-2=7


∴x³+1/x³
=(x+1/x)(x²-1+1/x²)
=3×(7-1)
=18

∵x^2-3x=1,显然x≠0,两边同时除以x得:x-3=1/x
∴x-1/x=3
∴(x-1/x)^2=9
∴x^2+1/x^2-2=9
∴x^2+1/x^2+2=13
即:(x+1/x)^2=13
∴x^3+1/x^3=(x+1/x)(x^2+1/x^2-1)=(x+1/x)[(x+1/x)^2-3]
=根号13×[13-3]=10根号13

x^2-3x+1=(x-3/2)^2-5/4=0
即(x-3/2)^2=5/4
解得x1=(3-根号5)/2 x2=(3+根号5)/2
把x的值代入式子就行了。