若√a-1+√ab-2=0,求1/ab+1/(a+1)(b+1)+1/(a+2)(b+2)+…+1/(a+2012)(b+2012)的值.
来源:学生作业帮助网 编辑:作业帮 时间:2024/09/06 19:52:07
![若√a-1+√ab-2=0,求1/ab+1/(a+1)(b+1)+1/(a+2)(b+2)+…+1/(a+2012)(b+2012)的值.](/uploads/image/z/8895682-10-2.jpg?t=%E8%8B%A5%E2%88%9Aa-1%2B%E2%88%9Aab-2%3D0%2C%E6%B1%821%2Fab%2B1%2F%28a%2B1%29%28b%2B1%29%2B1%2F%28a%2B2%29%28b%2B2%29%2B%E2%80%A6%2B1%2F%28a%2B2012%29%28b%2B2012%29%E7%9A%84%E5%80%BC.)
xQK
0!ژK]i/ t-yW{/Zt+of2o2bV֟&Ғ`Q+cVZ'@fb5Q*ȳcZ`VL,2@-XiEDXx ux$pgr?8Sm \Adjqx!3l^PrwQ
若√a-1+√ab-2=0,求1/ab+1/(a+1)(b+1)+1/(a+2)(b+2)+…+1/(a+2012)(b+2012)的值.
若√a-1+√ab-2=0,求1/ab+1/(a+1)(b+1)+1/(a+2)(b+2)+…+1/(a+2012)(b+2012)的值.
若√a-1+√ab-2=0,求1/ab+1/(a+1)(b+1)+1/(a+2)(b+2)+…+1/(a+2012)(b+2012)的值.
√(a-1)+√(ab-2)=0
√(a-1)=√(ab-2)=0
a-1=0,a=1
ab-2=0,b=2
1/ab+1/(a+1)(b+1)+1/(a+2)(b+2)+……+1/(a+2012)(b+2012)
=1/1×2+1/2×3+1/3×4+……+1/2013×2014
=1/1-1/2+1/2-1/3+1/3-1/4+……+1/2013-1/2014
=1-1/2014
=2013/2014
跪求u=u2(x) u2(y) u2(z)a(a2-2ab-b2)-b(2a2 ab-b2)∴2ab/a b≤2ab/2√ab=√ab |x-1| |x-2|>0
跪求u=u2(x) u2(y) u2(z)a(a2-2ab-b2)-b(2a2 ab-b2)|x-1| |x-2|>0∴2ab/a b≤2ab/2√ab=√ab
已知,√a-1+(ab-2)²=0,求
关于高中基本不等式若正数A,B满足AB=A+B+3,则AB的取值范围是:AB=A+B+3≥2√AB+3AB-2√AB-3≥0(√AB-3)(√AB+1)≥0(√AB)≥3AB)≥9(当且仅当A=B时取等号)以上没有问题疑惑:利用基本不等式 ab=0ab>=9或ab
已知a-b=,ab=1/8,求-2ab+ab+ab的值
已知(a-2)+(b+1)=0,求多项式3ab+ab-3ab+5ab+ab+2分之1ab
若ab^2+1=0,用因式分解法求-ab(a^2b^5-ab^3-b)
若ab^2+1=0,用因式分解法求-ab(a^2b^5-ab^3-b)
若a=√5+1.b=√5-1,求ab+ab
若a+b=1,ab≠0,求a/|a|+b/|b|+ab/|ab|的值
已知C是线段AB上的一点,BC=(√5-1)/2 AB,.若AB=a,求AC的长
已知(a+2)+|b+1|=0,求5ab-{2ab-[3ab-﹙4ab-2ab﹚]}的值
若a-b=4,ab=1,求(-2ab+2a+3b)-(3ab+2b-2a)-(a+4b+ab)的值
若a-b=4 ab=1 求(-2ab+2a+3b)-(3ab+2b-2a)-(a+4b+ab)的值
若a-b=4 ab=1 求(-2ab+2a+3b)-(3ab+2b-2a)-(a+4b+ab)的值
已知(a+b)的平方+|2b-1|=0,求6ab-2ab-3(ab-a)
若a+b+ab=1(a>0,b>0),求ab的最大值
已知|a-2|+|b+1|=0,求5a²bb-2ab+3ab