(a2-a)2-(a-1)2
来源:学生作业帮助网 编辑:作业帮 时间:2024/05/02 06:42:32
a^2+(a+1)^2+(a^2+a)^2=a^2+(a+1)^2+a^2(a+1)^2=a^2+(a+1)^2+2a(a+1)-2a(a+1)+a^2(a+1)^2=(a+1-a)^2+(a2+a)
2a-3b/b2-a2-a+3b/a2-b2+a+2b/a2-b2=(-2a+3b-a+3b+a+2b)/(a^2-b^2)=(-2a+8b)/(a^2-b^2)=-2(a-4b)/(a^2-b^2)
-√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)]
(a²-a-6)/(a+2)-√(a²-2a+1)/(a²-a)=(a-3)(a+2)/(a+2)-(a-1)/[a(a-1)]=a-3-1/a=1/2+√3-3-1/(
∵a2-3a+1=0,∴a2-3a=-1,a+1a=3,1+a2=3a,∴4a2-9a-2+91+a2,=4(a2-3a)+93a+3a-2,=4×(-1)+3(1a+a)-2,=-4+3×3-2,=
1/(a²-2a+1)-1/(a²+a-2)=1/(a-1)²-1/[(a+2)(a-1)]=[(a+2)-(a-1)]/[(a-1)²(a+2)]=3/[(a
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
能不能加上点括号让我们看明白到底谁是分子谁是分母?
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
(a-1)-(3a²-2a+1)=a-1-3a²+2a-1=-3a²+3a-2A=2(2-x)+1=4-2x+1=5-2x代入A-2b=x-15-2x-2b=x-12b=
(a2+a+1)(a2+a+2)-12=(a²+a)²+2(a²+a)+(a²+a)+2-12=(a²+a﹚²+3(a²+a)-1
(a^2+5a+2+1)(a^2+5a-2)-6=(a^2+5a+2)(a^2+5a-2)+a^2+5a-2-6=(a^2+5a)^2-4+a^2+5a-8=(a^2+5a)^2+a^2+5a-12=
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
(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=5a2-[a2+5a2-2a-2a2+6a]=5a2-[4a2+4a]=5a2-4a2-4a=a2-4a.
(1)∵A=-a-1,B=a2+a,a≠-1,∴B-A=(a2+a)-(-a-1)=a2+a+a+1=a2+2a+1=(a+1)2>0;(2)∵A=-a-1,C=2a2-5a-1,∴C-A=(2a2-
(a2-a)a2-2a-3÷a(a-3)=a(a-1)(a+1)(a-3)×a-3a=a-1a+1当a=2+1时,原式=22+1+1=22+2=2-1.故填空答案:2-1.