先化简在求值(a 2 a^2-2a 1-a a^2-4a 4
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![先化简在求值(a 2 a^2-2a 1-a a^2-4a 4](/uploads/image/f/2040042-66-2.jpg?t=%E5%85%88%E5%8C%96%E7%AE%80%E5%9C%A8%E6%B1%82%E5%80%BC%28a+2+a%5E2-2a+1-a+a%5E2-4a+4)
a/a+1-a平方-3a/a+3除以a平方-6a+9/a平方-9,其中a=-2=a/(a+1)-a(a+3)/(a+3)÷(a-3)²/(a-3)(a+3)=a/(a+1)-a÷(a-3)/
3(a+1)^2-(a+1)(2a-1)=3a^2+6a+3-2a^2-a+1=a^2+5a+4=3-5√3+4=7-5√3
你的题目估计抄错了!应该是:2a﹙a+b﹚-﹙a+b﹚²=﹙a+b﹚﹙2a-a-b﹚=﹙a+b﹚﹙a-b﹚=a²-b²=﹙√2011﹚²-﹙√2010﹚
a=1/(2+√3)=2-√3原式=(a+1)(a-1)/(a-1)-√(a-1)²/[a(a-1)]=a+1+(a-1)/[a(a-1)]=a+1+1/a=2-√3+1+1/(2-√3)=
:(a²+2a+1)/(a²-1)-a/(a-1)=(a+1)²/[(a+1)(a-1)]-a/(a-1)=(a+1)/(a-1)-a/(a-1)=(a+1-a)/(a-
原式=(a−2a(a+2)-a−1(a+2)2)•a+2a−4 =a2−4−a2+aa(a+2)2•a+2a−4 =1a2+2a.由a2+2a-1=0,得a2+2a=1,∴原式=1
[(a-2)/(a+3)]/[(a^2-4)/(2*a-6)]-[5/(a+2)]化简后等于-3/(a+2),这样就行了吧,可以带入值了
(a+b)(a-b)+a(2b-a)=a²-b²+2ab-a²=2ab-b²∴当a=1.5,b=2时,原式=2×1.5×2-2²=6-4=2
[(a^2-5a+2)/(a+2)+1]÷(a^2-4)/(a^2+4a+4)=(a^2-4a+4)/(a+2)÷(a-2)(a+2)/(a+2)^2=(a-2)^2/(a+2)÷(a-2)/(a+2
(2a-a^2)(a^2+4a+3)=a(2-a)(a+3)(a+1)(a^2+a)(a^2-5a+6)=a(a+1)(a-2)(a-3)所以[(2a-a^2)(a^2+4a+3)]/[(a^2+a)
原式=[a-2a(a+2)-a-1(a+2)2]•a+2a-4=a2-4-a2+aa(a+2)2•a+2a-4=a-4a(a+2)2•a+2a-4=1a(a+2).当a=2-1时,原式=1(2-1)(
a/(a-b)-a^2/(a^2-2ab+b^2)÷[(a/a+b-a^2/(a^2-b^2)]+1=a(a-b)/(a-b)^2-a^2/(a-b)^2÷[a(a-b)/(a^2-b)-a^2/(a
原式=(a²-6a+8)-(a²-4a+3)=a²-6a+8-a²+4a-3=-2a+5=-2×(-5/2)+5=5+5=10
原式=a2a-1-(a+1)=a2a-1-(a+1)(a-1)a-1=a2-a2+1a-1=1a-1,故答案为:1a-1.
a²-1/a-1×根号(a²-2a+1)/a²-a,=(a+1)*|a-1|/[a(a-1)]a>1时=1+1/aa1所以原式=1+1/a=(1+3*根号3)/(1+2*
a-(3b-a)-2b=a-3b+a-2b=2a-5b,a=2,b=-1/2时原式=4+5/2=13/2.
原式=a²+2ab-a²-2a-1+2a=2ab-1=2×2×(1/2)-1=2-1=1
-a²b+(3ab²-a²b)-2(2ab²-a²b)=-a²b+3ab²-a²b-4ab²+2a²