Nếu như vậy thì bạn hỏi cái gì ???????
Nếu như vậy thì bạn hỏi cái gì ???????
Tìm n biết:
8A=9-\(\frac{1}{3^n}\)và A=1+\(\frac{1}{3^2}+\frac{1}{3^4}+....+\frac{1}{3^{100}}\)
cho A = \(1+\frac{1}{3^2}+\frac{1}{3^4}+...+\frac{1}{3^{100}}\)
Biết 8A=9-\(\frac{1}{3^n}\)
Tính A = \(\frac{M}{N}\)biết
M =\(\frac{\frac{1}{99}+\frac{2}{98}+\frac{3}{97}+...+\frac{99}{1}}{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{100}}\)
N = \(N=\frac{92-\frac{1}{9}-\frac{2}{10}-\frac{3}{11}-...-\frac{90}{98}-\frac{91}{99}-\frac{92}{100}}{\frac{1}{45}+\frac{1}{50}+\frac{1}{55}+...+\frac{1}{495}+\frac{1}{500}}\)
1.Tìm x , y , z biết:
x : y : z = 3 : 4 : 5 và 2x^2 + 2y^3 - 3z^2 = -100
2.Tìm a^1 ; a^2 ; ...; a^9 biết:
\(\frac{a^1-1}{9}=\frac{a^2+2}{8}=\frac{a^3-3}{7}=...=\frac{a^9-9}{1}\)và a^1 + a^2 +...+a^9 = 90
3,Cho a/m = b/n = c/p = 4
Tính \(\frac{a+b+c}{m+n+p}\)
\(\frac{a-3b+2c}{m-3n+2p}\)
CMR
a)A=\(\frac{1}{3}+\frac{2}{3^2}+\frac{3}{3^3}+....+\frac{100}{3^{100}}< \frac{3}{4}\)
b)B=\(\frac{1}{3}+\frac{2}{3^2}+\frac{3}{3^3}+.....+\frac{100}{4^{100}}< \frac{4}{9}\)
a)A=\(\left(1-\frac{1}{1+2}\right)\left(1-\frac{1}{1+2+3}\right)...\left(1-\frac{1}{1+2+3+...+2006}\right)\)
b)\(B=1+\frac{3}{2^3}+\frac{4}{2^4}+...+\frac{100}{2^{100}}\)
c)C=\(\frac{1}{2!}+\frac{2}{3!}+...+\frac{n-1}{n!}\)
d)D=\(1+2^2+3^2+...+98^2\)
e)E=\(3^{100}-3^{99}+3^{98}-3^{97}+...+3^2-3+1\)
f)F=\(2^{2010}-2^{2009}-...-2-1\)
g)G=\(\left(\frac{1}{4}-1\right)\left(\frac{1}{9}-1\right)\left(\frac{1}{16}-1\right)...\left(\frac{1}{100}-1\right)\left(\frac{1}{121}-1\right)\)
Tinh a) \(\frac{\left(1+2+3+....+100\right).\left(\frac{1}{3}+\frac{1}{5}+\frac{1}{7}+\frac{1}{9}\right)}{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+..............+\frac{1}{100}}\)
Cho \(M=\frac{\frac{1}{99}+\frac{2}{98}+\frac{3}{97}+.......+\frac{99}{1}}{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}....+\frac{1}{100}}\)
\(N=\frac{92-\frac{1}{9}-\frac{2}{10}-\frac{3}{11}-....-\frac{92}{100}}{\frac{1}{45}+\frac{1}{50}+\frac{1}{55}+.....+\frac{1}{495}+\frac{1}{500}}\)
Tính M; N
CMR :
a , A = \(\frac{3}{1^2.2^2}+\frac{5}{2^2.3^2}+.....+\frac{19}{9^2.10^2}< 1\)
b , B = \(\frac{1}{3}+\frac{2}{3^2}+\frac{3}{3^3}+......+\frac{100}{3^{100}}< \frac{3}{4}\)
c, C = \(\left(\frac{1}{2^2}-1\right)\left(\frac{1}{3^2}-1\right).....\left(\frac{1}{100^2}-1\right)< \frac{1}{2}\)