3a[2a2-4a-3]Y-2A2[3A-4],QIZHONGA=-2

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已知A=a2-2a+5,B=a2-4a-3,试比较A、B大小.

∵A=a2-2a+5,B=a2-4a-3,∴A-B=(a2-2a+5)-(a2-4a-3)=a2-2a+5-a2+4a+3=2a+8,(1)当2a+8<0,即a<-4时,A<B;(2)当2a+8=0,

2a-3b/b2-a2 -a+3b/a2-b2 +a+2b/a2-b2

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)

设a=根号3-3,求[(a+1/a2-a)+(4/1-a2)]除以(a2+2a-3/a2+3a)的值

-√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)+

(a2-4/a2-4a+3)×(a-3/a2+3a+2=?

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)]

已知a2-3a+1=0,那么4a2−9a−2+91+a2=(  )

∵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,=

先化简,再求值:(4a2-3a)+(2+4a-a2)-(2a2+a-1),其中a=-2.

原式=4a2-3a+2+4a-a2-2a2-a+1=a2+3,当a=-2时,原式=4+3=7.

已知实数a满足a2+2a-1=0求(1 /a+1)-(a+3/a2-1)*(a2-2a+1/a2+4a+3)的值

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)-(

[1-(6+3a)/(a2+4a+4)]÷[(4a-4)/(a2+2a)]

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)=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=

已知a2-2a=-1,求3-2a2+4a的值.

a2-2a=-1,方程两边都乘以-2,得-2a2+4a=-2,方程两边都加3,得3-2a2+4a=3+(-2)=1.

已知a2+2a+1=0,求2a2+4a-3的值.

∵a2+2a+1=0,∴2a2+4a-3=2(a2+2a+1)-5=0-5=-5.

(a2+5a+3)(a2+5a-2)-6因式分解

(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=

先化简再求值:[(a+2)÷a2]÷(a-a分之4),a=根3

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√

(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)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

已知点P(a+1,3-a)点Q(2a,b+3)关于y轴对称,求代数式(a2+a)-(a2-b)的值

∵点P(a+1,3-a)点Q(2a,b+3)关于y轴对称,∴{a+1=-2a3-a=b+3解得:{a=-1/3b=1/3∴(a²+a)-(a²-b)=a²+a-a

a3-2a2-4a+3因式分解

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+

化简:5a2-[a2+(5a2-2a)-2(a2-3a)].

原式=5a2-[a2+5a2-2a-2a2+6a]=5a2-[4a2+4a]=5a2-4a2-4a=a2-4a.

(a2-a)a2-2a-3÷a(a-3)=a(a-1)(a+1)(a

(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.