Question

初二数学计算题一道

Answer 1

原式={(a+2)/[a(a-2)] -8/[(a+2)(a-2)]} × a/(a-2)
={[(a+2)²-8a]/[a(a-2)(a+2)]}× a/(a-2)
={(a²-4a+4)/[a(a-2)(a+2)]}× a/(a-2)
={(a-2)²/[(a-2)(a+2)]}× 1/(a-2)
=1/(a+2)

Answer 2

[(a+2)/(a²-2a)+8/(4-a²)]÷(a-2)/a
=[(a+2)/a(a-2)+8/(2-a)(2+a)】×a/(a-2)
=[(a²+4a+4)/a(a-2)(a+2)-8a/a(a-2)(a+2)]×a/（a-2)
=(a²+4a+4-8a)/a(a-2)(a+2)×a/(a-2)
=(a-2)²/a(a-2)(a+2)×a/(a-2)
=(a-2)/(a-2)(a+2)
=1/(a+2)

Answer 3

[(a+2)/(a^2-2a)+8/(4-a^2)]/[(a-2)/a]
=[(a+2)/(a(a-2))-8/((a+2)(a-2))]*[a/(a-2)]
=[(a+2)^2-8a]/[a(a+2)(a-2)]*[a/(a-2)]
=[a^2+4a+4-8a]/[a(a+2)(a-2)]*[a/(a-2)]
=[(a-2)^2]/[(a+2)(a-2)]*[1/(a-2)]
=1/(a+2)

Answer 4

解：原式=（a+2）/ (a-2)² -8a/(a-2)²（a+2）
=（a+2）²/ (a-2)²（a+2） -8a/(a-2)²（a+2）
=（a²+4a+4-8a）/(a-2)²（a+2）
=（a²-4a+4）/(a-2)²（a+2）
= (a-2) ² / (a-2)²（a+2）
= 1/（a+2）

Answer 5

（）里的先通分，得：[（a+2）^2 - 8a] / a(a+2)(a-2)
最后在除以右边的
得：1/(a+2)

Answer 6

原式=（（a+2)/a(a-2))+8/(2+a)(2-a)*a/(a-2)
=(a^2-4a+4)/a(a+2)(a-2)*a/(a-2)
=(a-2)^2/a(a+2)(a-2)*a/(a-2)
=(a-2)^2/(a+2)(a-2)^2
=1/(a+2)
分号，次方，实在不会打进去。将就看吧。。。。