求(1 a^2)-1 (1 a)通分式子及结果
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a-1/(a-1)(a+3),1-a/2(a-1)²=-1/2(a-1)先约分后再通分
在做题之前,我们首先要仔细观察两个分式的分母并进行比较分析,此题两分母的通分需要我们熟练掌握平方差公式.3a/2(a+1)=3a(a-1)/2(a+1)(a-1)…………分子分母同乘以(a-1)2a/
1/a^3-3a^2+2a=1/a(a-1)(a-2)=a(a+1)(a+2)/[a²(a²-1)(a²-4)],2/a^4-a^2=2/a²(a-1)(a+1
2a/(2a+1)=2a(2a-1)/[(2a-1)(2a+1)]4(2a-1)/(4a^2-4a+1)=4(2a-1)/(2a-1)²=4/(2a-1)=4(2a+1)/[(2a-1)(2
1/(a+b)=a(a-b)/[a(a+b)(a-b)]=a(a-b)/[a(a²-b²)]=(a²-ab)/[a(a²-b²)]1/(a²
1/(a²+a)与1/(a²-1)通分,公分母是(a³-a)1/(a²+a)--------------------(a²+a)是分母=1/[a(a+
解a-b=(a-b)(a²-b²)/(a²-b²)=(a-b)²(a+b)/(a²-b²)b/(a-b)=b(a+b)/(a
9-3a=-3(a-3)a²-9=(a+3)(a-3)分母(9-3a)与(a²-9)的最简公分母是3(a+3)(a-3)2/(9-3a)=2/[-3(a-3)]=-2/[3(a-3
a²+2a+1=(a+1)²所以最简公分母是a²(a+1)²所以(a-1)/(a²+2a+1)=a²(a-1)/[a²(a+1)&
a-1/a²+2a+1=(a-1)/(a+1)²=(a-1)²/[(a-1)(a+1)²]6/a²-1=6/[(a+1)(a-1)]=6(a+1)/[
答:a+2=(a+2)(2-a)/(2-a)=(4-a²)/(2-a)1/(2-a)
(a+1)/(a²-2a+1)=(a+1)/(a-1)²=(a+1)(1+a)/(1+a)(1-a)(1-a)6/(1-a²)=6/(1+a)(1-a)=6(1-a)/(
稍等,我看看再问:第二个分式有a-1不能约掉吗再答:不能约分。不然的话每题都能约分的。通分是为了让分母相同,从而可以进行分式的加减运算
a+2=(2+a)*(2-a)^2/(2-a)^24/(2-a)=4(2-a)/(2-a)^21/(2-a)^2
c/(a-b)=c(a-b)/(a-b)²1/(b-a)^2=1/(a-b)²
9-3a=-3(a-3)a²-9=(a+3)(a-3)所以2/(9-3a)=-(2a+6)/a(a+3)(a-3)=(-2a-6)/(a³-9a²)(a-1)/(a
=(a+1)(a-1)/(a+1)^2=(a-1)/(a+1)再问:对不起是a^2-a/a^2+2a+1再答:【补充1】(a^2-a)/(a^2+2a+1)=a(a-1)/(a+1)^2【补充2】(a
(a-1)/(a²+2a+1)和6/(a²-1)的分母分别是:a²+2a+1和a²-1.∵a²+2a+1=(a+1)²a²-1=(