化简a²-4分之2a-4 a 2分之2a 1
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∵a2(a2-2)-2a2+4=a4-2a2-2a2+4=a4-4a2+4=(a2-2)2≥0,又∵a为有理数,∴a2不等于2,∴整式≠0∴整式a2(a2-2)-2a2+4的值恒为正数.故选B.
-√3-2再问:能否写一下过程呢???再答:[(a+1)/(a²-a)+4/(1-a²)]/[(a²+2a-3)/(a²+3a)]=[(a+1)/a(a-1)+
原式=[(a²-4)/(a²-4a+3)]×[(a-3)/(a²+3a+2)]={(a-2)(a+2)/[(a-1)(a-3)]}×{(a-3)/[(a+1)(a+2)]
根据4+2a的大于0、小于0分别讨论当4+2a>=0,即a>=-2,原式=4+2a/根号a2+1当4+2a
1/(a+1)-(a+3)/(a^2-1)*(a^2-2a+1)/a^2+4a+3)=1/(a+1)-(a+3)/[(a-1)(a+1)]*(a-1)^2/[(a+1)(a+3)]=1/(a+1)-(
原式=(a²+4a+4-6-3a)/(a²+4a+4)×(a²+2a)/(4a-4)=(a+2)(a-1)/(a+2)²×a(a+2)/4(a-1)=a/4
代入X=12,则a=3,-a2/4-a=-21/4
具体步骤应为:原式=(a²-5a+2/a+2+a+2/a+2)÷[(a+2)(a-2)/(a+2)²]=(a²-4a+4/a+2)×(a+2)²/(a+2)(a
原式=[(a²-1)/(a²-2a+1)]+[(2a-a²)/(a-2)]÷a=[(a+1)(a-1)/(a-1)²]+[-a(a-2)/(a-2)]×(1/a
解原式=[(a+2)/a²]÷[(a²-4)/a]=(a+2)/a²×a/(a-2)(a+2)=1/a(a-2)=1/(√3)(√3-2)=1/(3-2√3)=(3+2√
再问:先化简再求值:[(a+2)÷a2]÷(a-a分之4),a=根3再答:
(a2-1)/(a2+2a+1)除以(a2-a)/(a+1)(a-2/(a+3)除以(a2-4)/(a2+6a+9)=﹙a²-1﹚/﹙a²+2a+1﹚×﹙a+1﹚/﹙a²
原式==[(a-b)/(a+2b)]÷[(a+b)(a-b)/(a+2b)²]=(a-b)/(a+2b)×(a+2b)²/(a+b)(a-b)=(a+2b)/(a+b)
a³-2a²-4a+3=a³-2a²-3a-a+3=a(a²-2a-3)-(a-3)=a(a+1)(a-3)-(a-3)=(a-3)(a²+
a2-1分之a乘a2-a负二方可以化简为①(a²-1)/a乘(a²-a²/1)≤4②(a²-1)/a乘a²/(a²-1)(a²+1
不好意思再答:看错了再问:,,提负号有问题吧,,再答:对的呀再问:谢了再答:不客气
①a2-4a+4=(a-2)2,符合;②a2+a+14=(a+12)2,符合;③4a2-a+14,不符合;④4a2+4a+1=(2a+1)2,符合.故应填:①②④.
(a-1)/(a²+2a+1)和6/(a²-1)的分母分别是:a²+2a+1和a²-1.∵a²+2a+1=(a+1)²a²-1=(
原式=5a2-[a2+5a2-2a-2a2+6a]=5a2-[4a2+4a]=5a2-4a2-4a=a2-4a.