先化简,再求值a分之a-4÷(a的平方-2a分之 2)
来源:学生作业帮助网 编辑:作业帮 时间:2024/04/30 08:57:42
[(a-2)/a(a+2)-(a-1)/(a+2)^2]除以((a-4)/(a+2)=(a^2-4-a^2+a)/a(a+2)^2*(a+2)/(a-4)=(a-4)/a(a+2)^2*(a+2)/(
原式=[a/(a+2)+2/(a-2)]÷1/(a-4)=[a*(a-2)+2*(a+2)]/(a+2)*(a-2)÷1/(a+2)*(a-2)=a*(a-2)+2*(a+2)=2a另取a=1,则由2
(a-1/a)=(a²-1)/a=.(a+1)(a-1)/a÷a分之a方+2a+1=÷(a²+2a+1)a=÷(a+1)(a+1)/a=*a/[(a+1)(a+1)](a-a分之1
=[(a²-4)/(a-2)]×1/a(a+2)=[(a+2)(a-2)/(a-2)]×1/a(a+2)=(a+2)×1/a(a+2)=1/a
4/(a^2-4)+2/(a+2)-1/(a-2)=4/(a+2)(a-2)+2(a-2)/(a+2)(a-2)-(a+2)/(a+2)(a-2)=[4+2a-4-a-2]/(a+2)(a-2)=(a
=(a-3)/2(a-2)÷[5/(a-2)-(a+2)(a-2)/(a-2)]=(a-3)/2(a-2)÷(5-a²+4)/(a-2)=(a-3)/2(a-2)×[-(a-2)/(a+3)
[(a+2)分之a+(a-2)分之2]÷(a²-4)分之1=[(a+2)分之a+(a-2)分之2]×(a²-4)=(a+2)分之a×(a²-4)+(a-2)分之2×(a&
(a²+a)(a²-5a+6)分之(2a-a²)(a²+4a+3)=(2a-a²)(a²+4a+3)/(a²+a)(a²
a²-ab分之b²-a²÷【a+a分之2ab+b²】×【a分之1+b分之1】,=-(a+b)(a-b)/a(a-b)×a/(a+b)²×(a+b)/a
2a-4分之3-a/(a+2-a-2分之5),=(3-a)/2(a-2)÷[(a+2)(a-2)-5]/(a-2)=(3-a)/2÷(a²-9)=(3-a)/2÷[(a+3)(a-3)]=-
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=(a²+ab-a²)/(a+b)×(a-a+b)/(a-b)×(a+b)/b²=ab/(a+b)×b/(a-b)×(a+b)/b²=a/(a-b)
(a+2a+1)/(a-1)-a/(a-1)=(a+1)^2/(a+1)(a-1)-a/(a-1)=(a+1)/(a-1)-a/(a-1)=1/(a-1)由a=根号3+1,所以原式=1/(a-1)=1
原式=(3+a)(3-a)/(a+3)^2x(a+3)/(a-3)a-1/a=-1/a-1/a=-2/a=-2/根号2=-根号2
(a分之a^2+1-2)÷a^2+2a分之(a+2)(a-1),因为a^2=4,所以a=2或-2,又a+2若为0,即a=-2,公式分母为0,公式无意义,所以a=2,然后化简得结果为a-2,其中a=2,
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√
http://ci.baidu.com/ALnuXPtkhX
(a-2)分之a-(a²-2a)分之2a-a分之2=[a(a-2)]分之a²-[a(a-2)]分之(2a)-[a(a-2)]分之[2(a-2)]=[a(a-2)]分之[a²