先化简再求值,a+1除以a的平方-搜答案-智慧作业帮

[(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)/(

[(a-2)/(a²+2a)-(a-1)/(a²+4a+4)]÷(a-4)/(a+2)=(a-2)/(a(a+2))*(a+2)/(a-4)-(a-1)/(a+2)²*(

答：题目最后是tan6?没写全吧?是不是tan60°?[2/(a+1)+(a+2)/(a²-1)]÷[a/(a+1)]=[2(a-1)+(a+2)]/[(a+1)(a-1)]×(a+1)/a

(a+1分之1-a的平方-1分之a-2)除以a+1分之1=1/(a+1)x(a+1)-(a-2)/(a+1)(a-1)x(a+1)=1-(a-2)/(a+1)=(a+1-a+1)/(a+1)=2/(a

原式=a+(1-a)的绝对值∵1-a＜0∴原式=a+(a-1)=2a-1当a=9时原式=2a-1=2×9-1=17

原式=(a+b)(a-b)/ab(a-b)÷(a+b)²/2ab=(a+b)/ab×2ab/(a+b)²=2/(a+b)∴原式=2/(5-√11-3+√11)=1

(a+1)的平方-(3a的平方+a)除以a=a²+2a+1-3a-1=a²-a=9+3=12

（x-2）²-（x+2）（x-1),=x^2-4x+4-x^2-x+2=-5x+5=0a²*a的4次方-a的8次方除以a²+（a的3次方）²,=a^6-a^6+

原式=(a-2)/(a-4)÷(a²-4a+4)/(a-4)=(a-2)/(a-4)×(a-4)/(a-2)²=1/(a-2)=1/(√3+2-2)=√3/3

[1/(a-1)+a/(1-a)]/a=[1/(a-1)-a/(a-1)]/a=[(1-a)/(a-1)]/a=-1/a把a=√2+1代入-1/(√2+1)=-1(√2-1)/(√2+1)(√2-1)

第一题：两个方程相加：3y-2=3/4y=11/12;将y值代入方程就可求得x=12/31第二题：第一个方程x3得：9/X-3/y=15将新得方程与第二方程相加得11/x=16x=11/16将x值代入

能不能把原式再确认一下?感觉有问题.

不用化简,a=1直接代入计算也方便.(a-3)/(2a-4)除以(5/(a-2)-a-2)=(1-3)/(2-4)÷[5/(1-2)-1-2]=1÷(-5-3)=-1/8再问：帮我化一下，我的老师狡猾

[1-1/(a-1)]÷[(a²-4a+4)/(a²-a)]=(a-2)/(a-1)÷[(a-2)²/a(a-1)]=(a-2)/(a-1)*[a(a-1)/(a-2)&

原式=(a-2)/(a+1)(a-1)÷[(a-1-2a+3)/(a-1)]=(a-2)/(a-1)(a+1)÷(2-a)/(a-1)=(a-2)/(a+1)(a-1)x(a-1)...