第23页

信息发布者:
$解:(1)解:\frac{x^2}{x-1}=\frac{x^2-1+1}{x-1}=x+1+\frac{1}{x-1}.$
$(2)\frac{3x+3-5}{x+1}=3-\frac{5}{x+1},$
$∵x为整数,分式\frac{3x-2}{x+1}的值为整数,$
$∴x+1=1,5,-1,-5,$
$∴x=0,4,-2,-6.$
B
二、三
$解:∵x+\frac{1}{x}=3,∴(x+\frac{1}{x})^2=9,即x^2+2+\frac{1}{x^2}=9.∴x^2+\frac{1}{x^2}=7.$
$∵(x+\frac 1x)^3=27,∴x^3+\frac{1}{x^3}+3(x+\frac{1}{x})=27.∴x^3+\frac{1}{x^3}=18.$
$∵x^4+\frac{1}{x^4}=(x^2+\frac{1}{x^2})^2-2=47,$
$∴原式=\frac{18+7}{47+3}=\frac{1}{2}.$