一道分式化简求值先化简,再代值[(x+2)/(x²-2x) – (x-1)/(x²-4x+4)]÷(x²-16)/(x²+4x)其中x=2+根2
问题描述：
一道分式化简求值
先化简,再代值
[(x+2)/(x²-2x) – (x-1)/(x²-4x+4)]÷(x²-16)/(x²+4x)
其中x=2+根2
答
代入得[(x+2)/(x²-2x) – (x-1)/(x²-4x+4)]÷(x²-16)/(x²+4x)
=[(x+2)/x(x-2) – (x-1)/(x-2)²]÷[(x+4)(x-4)/x(x+4)]
={[(x+2)(x-2)-x(x-1)]/x(x-2)²}÷[(x-4)/x]
=[(x²-4-x²+x)/x(x-2)²]×[x/(x-4)]
=[(x-4)/x(x-2)²]×[x/(x-4)]
=1/(x-2)²
=1/(2+√2-2)²
=1/2
答
[(x+2)/(x²-2x) – (x-1)/(x²-4x+4)]÷(x²-16)/(x²+4x)=[(x+2)/x(x-2) – (x-1)/(x-2)²]÷[(x+4)(x-4)/x(x+4)]={[(x+2)(x-2)-x(x-1)]/x(x-2)²}÷[(x-4)/x]=[(x²-4-x²+x)/x(...