
巧算:(1/x-1)-(1/x+1)-(2/x^2+1)-(4/x^2+1)-(8/x^2+1)
1个回答
展开全部
解:(1/x-1)-(1/x+1)-(2/x^2+1)-(4/x^2+1)-(8/x^2+1)
=(1/x-1)-(1/x+1)-(2+4+8)/(x^2+1)
=(x+1)/(x²-1)-(x-1)/(x²-1)-(2+4+8)/(x^2+1)
=(x+1-x+1)/(x²-1)-(14)/(x^2+1)
=2/(x²-1)-14/(x^2+1)
=2(x²+1)/(x²-1)(x²+1)-14(x²-1)/(x^2+1)(x²-1)
=(-12x²+16)/(x^4-1)
=(1/x-1)-(1/x+1)-(2+4+8)/(x^2+1)
=(x+1)/(x²-1)-(x-1)/(x²-1)-(2+4+8)/(x^2+1)
=(x+1-x+1)/(x²-1)-(14)/(x^2+1)
=2/(x²-1)-14/(x^2+1)
=2(x²+1)/(x²-1)(x²+1)-14(x²-1)/(x^2+1)(x²-1)
=(-12x²+16)/(x^4-1)
推荐律师服务:
若未解决您的问题,请您详细描述您的问题,通过百度律临进行免费专业咨询