Các câu hỏi tương tự
chứng minh rằng:
a) A= \(\frac{3}{1^2.2^2}+\frac{5}{2^2.3^2}+\frac{7}{3^2.4^2}+...+\frac{19}{9^2.10^2}\)<1
b)B=\(\frac{1}{3}+\frac{2}{3^2}+\frac{3}{3^3}+\frac{4}{4^4}+...+\frac{100}{3^{100}}< \frac{3}{4}\)
Chướng minh rằng:
a, \(\frac{1}{1^2.2^2}\)+$\frac{5}{2^2.3^2}$+$\frac{5}{3^2.4^2}$+...+$\frac{5}{9^2.10^2}$ <1
b, \(\frac{1}{3}\)+\(\frac{2}{3^2}\)+$\frac{3}{3^3}$+$\frac{4}{3^4}$+...+$\frac{100}{3^100}$ <\(\frac{3}{4}\)
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 = \(\left(\frac{1}{2^2}-1\right)\left(\frac{1}{3^2}-1\right).....\left(\frac{1}{100^2}-1\right)< \frac{1}{2}\)
Rút gọn: a)\(\frac{2^{19}.27^3+15.4^9.9^4}{6^9.2^{10}.12^{10}}\)
b)\(\frac{\left(-\frac{1}{2}\right)^3-\left(\frac{3}{4}\right)^3.\left(-2\right)^2}{2.\left(-1\right)^5+\left(\frac{3}{4}\right)^2-\frac{3}{8}}\)
Tìm x
a)\(3^{x+1}=9^x\)
b)\(2^{3x+2}=4^{x+5}\)
c)\(3^{2x-1}=243\)
tính A= \(\frac{3}{1^2.2^2}+\frac{5}{2^2.3^2}+\frac{7}{3^2.4^2}+....+\frac{19}{9^2.10^2}\)
giải hắn ra
Chướng minh rằng:
a, \(\frac{1}{1^2.2^2}\)+\(\frac{5}{2^2.3^2}\) +\(\frac{5}{3^2.4^2}+...+\frac{5}{9^2.10^2}\)<1
b, \(\frac{1}{3}\)+\(\frac{2}{3^2}\) +\(\frac{3}{3^3}+\frac{4}{3^4}+...+\frac{100}{3^{100}}\)<\(\frac{3}{4}\)
Tính :
\(\frac{3}{1^2.2^2}+\frac{5}{2^2.3^2}+\frac{7}{3^2.4^2}+....+\frac{19}{9^2.10^2}\)
27 tháng 1 2015 lúc 14:36
Tính tổng :\(\frac{3}{1^2.2^2}+\frac{5}{2^2.3^2}+\frac{7}{3^2.4^2}+........+\frac{19}{9^2.10^2}\)
A=\(\frac{1}{3}+\frac{2}{3^2}+\frac{3}{3^3}+...+\frac{100}{3^{100}}\)\(< \frac{3}{4}\)
B=\(\frac{3}{1^2.2^2}+\frac{5}{2^2.3^2}+\frac{7}{3^2.4^2}+...+\frac{19}{9^2.10^2}< 1\)
C=\(1+3+3^2+3^3+...+3^{100}\)
D=\(2^{100}-2^{99}+2^{98}-2^{97}+...+2^2-2\)