1.計(jì)算
1.計(jì)算
(x+2)/(x+1)-(x+3)/(x+2)-(x-4)/(x-3)+(x-5)/(x-4)
2.解方程
2x/(2x-1)+x/(x-2)=2
3.先化簡(jiǎn)再求值 求(1/x²-1/y²)[(x²-xy+y²)/(x-y)+(x²+xy+y²)/(x+y)]的值 ,其中x=1/2 y=1/4
(x+2)/(x+1)-(x+3)/(x+2)-(x-4)/(x-3)+(x-5)/(x-4)
2.解方程
2x/(2x-1)+x/(x-2)=2
3.先化簡(jiǎn)再求值 求(1/x²-1/y²)[(x²-xy+y²)/(x-y)+(x²+xy+y²)/(x+y)]的值 ,其中x=1/2 y=1/4
數(shù)學(xué)人氣：496 ℃時(shí)間：2020-10-01 19:23:46
優(yōu)質(zhì)解答
第一個(gè)：(x+2)/(x+1)-(x+3)/(x+2)-(x-4)/(x-3)+(x-5)/(x-4)=1+1/(x+1) - 1 - 1/(x+2) - 1 +1/(x-3) +1 -1/(x-4)= (x+2 -x -1)/[(x+1)*(x+2)] +(x-4 -x +3)/[(x-3)*(x-4)]=1/[(x+1)*(x+2)] +(-1)/[(x-3)*(x-4)]=[(x-3...
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