已知X+1/X=2,求分式X^2+X+1/(3X^2-5X+3)的值

问题描述:

已知X+1/X=2,求分式X^2+X+1/(3X^2-5X+3)的值

同时处以x,得出(X^2+X+1)/(3X^2-5X+3)=(x+1+1/x)/(3x-5+3/x)=(2+1)/(3x2-5)=3/1=3

X+1/X=2,(X-1)^2=0,X=1
X^2+X+1/(3X^2-5X+3)=3

x+1/x=2
X²+1=2X 或 x+1/x=1+1/1
X²-2X+1=0 x=1
(X-1)²=0
X=1
代入:
(1+1+1)/(3-5+3)=3
原式=3
你皋城初一的不?在做数学乐园?校友.

X+1/X=2
(X^2+1)/X=2
则 X^2=2X-1
X^2+X+1/(3X^2-5X+3)
=2X-1+X+1/(6X-3-5X+3)
=X-1+1/X
=X+1/X-1
=2-1
=1