来自孟昕元的问题
先化简再求值:(x分之-1-x-1分之x-2)除以x方+2x+1分之2x方+1,其中x满足x方-x-1=0
先化简再求值:(x分之-1-x-1分之x-2)除以x方+2x+1分之2x方+1,其中x满足x方-x-1=0
3回答
2020-09-24 22:05
先化简再求值:(x分之-1-x-1分之x-2)除以x方+2x+1分之2x方+1,其中x满足x方-x-1=0
先化简再求值:(x分之-1-x-1分之x-2)除以x方+2x+1分之2x方+1,其中x满足x方-x-1=0
原式=(-1/x-(x-2)/(x-1))÷2x^2/(x^2+2x+1)+1=[(-(x-1)-x(x-2)]/x(x-1)×(x+1)^2/2x^2+1=(-x+1-x^2+2x)/x(x-1)*(x+1)^2/2x^2+1=(-x^2+x+1)/(x^2-x)*(x+1)^2/2x^2+1∵x^2-x-1=0x^2=x+1x^2-x=1=(-1+1)/1*x^4/2x^2+1=0+...
解释一下
=(-1/x-(x-2)/(x-1))÷2x^2/(x^2+2x+1)+1
=[(-(x-1)-x(x-2)]/x(x-1)×(x+1)^2/2x^2+1
=(-x+1-x^2+2x)/x(x-1)*(x+1)^2/2x^2+1
=(-x^2+x+1)/(x^2-x)*(x+1)^2/2x^2+1
谢谢
第二步时前面是通分后面是除号变乘号要分母和分子倒位置
第三步是去括号,同项化简
第四步也是化简