a^2-7a 2
来源:学生作业帮助网 编辑:作业帮 时间:2024/07/09 16:47:52
![a^2-7a 2](/uploads/image/f/469536-24-6.jpg?t=a%5E2-7a+2)
∵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.
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)]
2011a^2-3a=22a(a^2-3a)+a^2-7a+2009=2a*2+a^2-7a+2009=a^2-3a+2009=2+2009=2011
∵代数式a2+a+3的值为7,∴a2+a+3=7,∴a2+a=4,∴2a2+2a-3=2(a2+a)-3=2×4-3=5.
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)-(
a2-2a-3a2-7a+12=(a-3)(a+1)(a-3)(a-4)=a+1a-4,当a=23时,原式=23+123-4=-12.
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=
(1)3a2-2a+4a2-7a=7a2-9a(2)-3a+[4b-(a-3b)]=-4a+7
(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=
(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
(1)由题意,知a3-2a2-a+7=5,解得a=-1,1,2.当a=-1时,A={2,4,5},B={-4,2,4,5},此时A∩B={2,4,5}与已知A∩B={2,5}矛盾;当a=1时 
您好很高兴为您解答疑难@!由a2-7a+2=0b2-7b+2=0得a2-7a=b2-7b=-2移项:a2-b2=7a-7b(a+b)*(a-b)=7(a-b)解得:a+b=72a2+b2-7a+1=a
(a²-b²)²-2(a²+b²)(a+b)²=[(a-b)(a+b)]²-2(a²+b²)(a+b)
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-