=[(a+2)/a(a-2) -(a-1)/(a-2)²]a/(4-a)
=[(a+2)(a-2)-a(a-1)]a/[(4-a)a(a-2)²]
=(a²-4-a²+a)/[(a-2)²(4-a)]
=(a-4)/[(a-2)²(4-a)]
=-1/(a-2)²
=（4-a)/a(a-2)÷[(a+2)/a(a-2)-(a-1)/(a-2)²]
=(4-a)/a(a-2)÷[(a²-4-a²+a)/a(a-2)²]
=(4-a)/a(a-2)÷[(a-4)/a(a-2)²]
=-1/(a-2)
=1/(2-a)
计算(A+2\/A^2-2A-A-1\/A^2-4A+4)*A-2\/A÷4-A\/A^2
【（A + 2）\/ （A ² - 2 A）- （A - 1）\/ （A ² - 4 A + 4）】 × （A - 2）\/ A ÷ （4 - A）\/ A ²= 【（A + 2）\/ A（A - 2）- （A - 1）\/ （A - 2）² 】× （A - 2）\/ A × A ² \/ （4 - A）= 【（A +...
...a-2\/a^2+2a - a-1\/a^2+4a+4)÷a-4\/a+2 其中a满足 a^2+2a-1=0...
[(a-2)\/(a^2+2a)-(a-1)\/(a^2+4a+4)]\/a-4\/a+2,其中a满足a^2+2a-1=0 因为a^2+2a-1=0, 所以a^2+2a=1, 所以[(a-2)\/(a^2+2a)-(a-1)\/(a^2+4a+4)]\/[(a-4)\/(a+2) =[(a-2)\/a(a+2)-(a-1)\/(a+2)^2]*[(a+2)\/(a-4)] =[(a-2)(a+2)\/....
(a+2)\/(a^2-2a)-(a-1)\/(a^2-4a+4)*(a^2-2a)\/(4-a)怎么做
原式=(a+2)\/a(a-2)-(a-1)\/(a-2)^2*a(a-2)\/(4-a)=(a+2)\/a(a-2)+a(a-1)\/(a-2)(a-4)=[(a+2)(a-4)+a^2(a-1)]\/a(a-2)(a-4)=(a^2-2a-8+a^3-a^2)\/a(a-2)(a-4)=(a^3-2a-8)\/a(a-2)(a-4)
...2\/(a^2+2a)-a-1\/(a^2+4a+4))÷a-4\/a+2,其中a=√2-1
我的 先化简,再求值:(a-2\/(a^2+2a)-a-1\/(a^2+4a+4))÷a-4\/a+2,其中a=√2-1 1个回答 #热议# 什么样的人容易遇上渣男?d1353568405 2013-03-09 · TA获得超过5001个赞 知道小有建树答主 回答量:1116 采纳率:100% 帮助的人:633万 我也去答题访问个人页 关注 展开全部 来自:...
化简,求值:(a-2\\a2+2a-a-1\\a2+4a+4)\\a-4\\a+2,其中a2+2a-1=0
所以a^2+2a=1,所以[(a-2)\/(a^2+2a)-(a-1)\/(a^2+4a+4)]\/[(a-4)\/(a+2)=[(a-2)\/a(a+2)-(a-1)\/(a+2)^2]*[(a+2)\/(a-4)]=[(a-2)(a+2)\/a(a+2)^2-a(a-1)\/a(a+2)^2]*[(a+2)\/(a-4)]=[(a^2-4-a^2+a)\/a(a+2)^2]*[(a+2)\/(a-4...
计算: a+2\/a^2-2a×a^2-4a+4\/a+2 , x-2\/x^2-1÷x+1\/x^2+2x+1 , (2x...
第1题应该是((a+2)\/(a^2-2a))*((a^2-4a+4)\/(a+2))吧!如果是就=((a+2)\/a(a-2))*((a-2)^2\/(a+2))约分得(a-2)\/a 2.原式=((x-2)\/(x+1)(x-1))*((x+1)^2\/(x+1))约分=(x-2)\/(x-1)3.(2x\/x-2-x\/x+2)\/x\/x^2-4 =2x\/(x-2x)-x\/(x+2)*...
先化简,在求值:(a-2\/a+2a - a-1\/a+4a+4)除于a-4\/a+2,其中a满足:a2+2a...
先把总式化之最简,为（6a2+4a-3）（a+2）\/(a2-4a),将条件中a2+2a=1代入,可将化简式中平方消去,从而再将a2+2a-1=0解得的根代入,即得最终值
化简,求值:(a-2\/a^2 2a-a-1\/a^2 4a 4)\/a-4\/a+2,其中a满足a^2+2a-1=0...
(a-2\/a^2 2a-a-1\/a^2 4a 4)\/a-4\/a+2 =[(a-2)\/a(a+2)-(a-1)\/(a+2)²]×(a+2)\/(a-4)=[a²-4-a²+a]\/a(a+2)²×(a+2)\/(a-4)=1\/a(a+2)=1\/(a²+2a)=1\/1 =1;请采纳 如果你认可我的回答，敬请及时采纳，~如果你认可我...
(a-2\/a2+2a-a-1\/a2+4a+4)\/a-4\/a+2,其中a满足a2+2a=1
(a-2)\/(a+2a)-(a-1)\/(a+4a+4)-(a-4)\/(a+2)...[猜着修改的] =(a-2)\/(a+2a)-(a-1)\/(a+4a+4)-(a-4)\/(a+2) =(a-2)\/a(a+2)-(a-1)\/(a+2)-(a-4)\/(a+2) =(a-4)\/a(a+2)-a(a-1)\/a(a+2)-(a-4)\/(a+2) =(a-4)\/a(a+2)-(a-4)\/(...
化简,求值:((a-2)\/2a∧2+(a-1)\/(a∧2+4a+4))\/(a-4)\/(a+4),其_百度知 ...
所以a^2+2a=1,所以[(a-2)\/(a^2+2a)-(a-1)\/(a^2+4a+4)]\/[(a-4)\/(a+2)=[(a-2)\/a(a+2)-(a-1)\/(a+2)^2]*[(a+2)\/(a-4)]=[(a-2)(a+2)\/a(a+2)^2-a(a-1)\/a(a+2)^2]*[(a+2)\/(a-4)]=[(a^2-4-a^2+a)\/a(a+2)^2]*[(a+2)\/(a-4...
