设a=根号3-3,求[(a+1/a2-a)+(4/1-a2)]除以(a2+2a-3/a2+3a)的值
来源:学生作业帮 编辑:作业帮 分类:综合作业 时间:2024/05/10 08:17:51
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)+4/(1-a)(1+a)]/[(a+3)(a-1)/a(a+3)] =[(a+1)/a(a-1)-4/(a-1)(a+1)]/[(a-1)/a] ={[(a+1)²/[a(a-1)(a+1)]-4a/[a(a-1)(a+1)]}/[(a-1)/a] =(a-1)²/[a(a-1)(a+1)]/[(a-1)/a] =(a-1)/[a(a+1)]/[(a-1)/a] =1/(a+1) =1/(√3-3+1) =1/(√3-2) =1(√3+2)/(√3-2)(√3+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)+4/(1-a)(1+a)]/[(a+3)(a-1)/a(a+3)] =[(a+1)/a(a-1)-4/(a-1)(a+1)]/[(a-1)/a] ={[(a+1)²/[a(a-1)(a+1)]-4a/[a(a-1)(a+1)]}/[(a-1)/a] =(a-1)²/[a(a-1)(a+1)]/[(a-1)/a] =(a-1)/[a(a+1)]/[(a-1)/a] =1/(a+1) =1/(√3-3+1) =1/(√3-2) =1(√3+2)/(√3-2)(√3+2) =-√3-2
设a=根号3-3,求[(a+1/a2-a)+(4/1-a2)]除以(a2+2a-3/a2+3a)的值
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
(a2-1)/(a2+2a+1)除以(a2-a)/(a+1) (a-2/(a+3)除以(a2-4)/(a2+6a+9)
已知a=1/2+根号3,求a2-a-6/a+2 - 根号a2-2a+1/a2-a的值
已知a2-2a=-1,求3-2a2+4a的值.
已知a2+2a+1=0,求2a2+4a-3的值.
若a2-3a+1=0,求a2+a2/1的值
已知a2+3a+1=0 求 1+1/a a2+1/a2
(a2-4/a2-4a+3)×(a-3/a2+3a+2=?
设a2+a-1=0,求2a3+4a2+1998的值.
已知a2+a+1=0,求a三次方+2a2+2a-3的值
若根号(a2-3a+1)+b2+2b+1,则a2+1/a2-绝对值b=是多少