化简求值 (1-a 1分之1)÷a² 2a 1分之a 其中a=根号3-1
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=1/(a+2)+4/(a+2)(a-2)=(a-2+4)/(a+2)(a-2)=1/(a-2)=1/(2+√3-2)=√3/3
原式=[(a+b)(a-b)/a(a-b)]÷[(a²+2ab+b²)/a]=[(a+b)/a]×[a/(a+b)²]=1/(a+b)b=-1假设a=2原式=1/(2-1
(a^2+2a+1)/(a^2-1)+1/(a-1)=(a+1)/(a-1)+1/(a-1)=(a+2)/(a-1)=(根号2+1+1)/(根号2-1+1)=根号2(根号2+2)/根号2²=
(a²-4a+3)分之(a²-4)×(a²+3a+2)分之(a-3)-(a²-1)分之(a+3)=(a+2)(a-2)/(a-3)(a-1)x(a-3)/(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
2分之1a-(2a-3分之b²)+(-2分之3a+3分之1b²),=2分之1a-2a+3分之b²-2分之3a+3分之1b²=-3a+﹙2/3﹚b²,把
1/a+√(1/a²+a²-2)=1/a+√(1/a-a)²=1/a+|1/a-a|∵1/a=5>a∴原式=1/a+1/a-a=2/a-a=10-1/5=49/5∴甲的对
a=根号2-1∴a-1
原式=1/a+√(1/a²+a²-2)=1/a+√(1/a-a)²=1/a+1/a-a=2/a-a=10-1/5=9.8
解1/(a+1)-(a+1)/(a²-2a+1)÷(a+1)/(a-1)=1/(a+1)-(a+1)/(a-1)²×(a-1)/(a+1)=1/(a+1)-1/(a-1)=(a-1
原式=(a+1)/(a-1)-[a/(1/a)]/(a²-2a+1)=(a+1)/(a-1)-a²/(a-1)²=[(a+1)(a-1)-a²]/(a-1)&s
(a²+2ab+b²)/(a²-b²)=(a+b)²/((a+b)(a-b))=(a+b)/(a-b)=(1/b+1/a)/(1/b-1/a)=(2根
原式=(2aa−1-aa−1)÷a=aa−1×1a=1a−1,当a=2+1时,原式=12+1−1=12=22.
加点括号吧,求你了再问:(a方+2a+1)分之(a方-1)+(a-2)分之(2a-a方)除以(a)其中a=(根号2)+1再答:不知你的“除以(a)”是啥意思?除在哪里?只能给你下式(a^2-1)/(a
(1)∵a12+a−12=3,∴两边平方,得a+a-1+2=9,∴a+a-1=7.把a+a-1=7两边平方,得a2+a-2+2=49,∴a2+a-2=47. (2)(lg5)2+lg2×lg
[(x+2)/x-(x-1)/(x-2)]÷(x-4)/x=[1+2/x-1-1/(x-2)]×x/(x-4)=(x-4)/x(x-2)×x/(x-4)=1/(x-2)=1/(3-2)=1这是我在静心
(a-1)²+2a(a-1)提取(a-1)得=(a-1)(a-1+2a)=(a-1)(3a-1)=(2-1)(6-1)=1*5=5