Đặt BT là A
\(\Rightarrow A=\frac{2^2-1^2}{1^2.2^2}+\frac{3^2-2^2}{2^2.3^2}+....+\frac{10^2-9^2}{9^2.10^2}\)
\(A=\frac{1}{1^2}-\frac{1}{2^2}+\frac{1}{3^2}-\frac{1}{2^2}+....+\frac{1}{9^2}-\frac{1}{10^2}\)
\(A=1-\frac{1}{10^2}< 1\)
=> A<1(đpcm)
Đặt BT là A
\(\Rightarrow A=\frac{2^2-1^2}{1^2.2^2}+\frac{3^2-2^2}{2^2.3^2}+....+\frac{10^2-9^2}{9^2.10^2}\)
\(A=\frac{1}{1^2}-\frac{1}{2^2}+\frac{1}{3^2}-\frac{1}{2^2}+....+\frac{1}{9^2}-\frac{1}{10^2}\)
\(A=1-\frac{1}{10^2}< 1\)
=> A<1(đpcm)
Chứng minh rằng: \(\frac{3}{1^2.2^2}+\frac{5}{2^2.3^2}+\frac{7}{3^2.4^2}+......+\frac{19}{9^2.10^2}< 1\)
Chứng minh rằng:
\(\frac{3}{1^2.2^2}+\frac{5}{2^2.3^2}+\frac{7}{3^2.4^2}+...+\frac{19}{9^2.10^2}< 1\)
1.Chứng minh rằng: \(\frac{3}{1^2.2^2}+\frac{5}{2^2.3^2}+\frac{7}{3^3.4^2}+...+\frac{19}{9^2.10^2}< 1\)
2.Chứng minh rằng: \(\frac{1}{3}+\frac{2}{3^2}+\frac{3}{3^3}+\frac{4}{3^4}+...+\frac{100}{3^{100}}< \frac{3}{4}\)
Làm nhanh giúp mình nhé mọi người !!!
Chứng minh:
\(\dfrac{3}{1^2.2^2}+\dfrac{5}{2^2.3^2}+\dfrac{7}{3^2.4^2}+...+\dfrac{19}{9^2.10^2}< 1\)
Cho \(M=\frac{3}{1^2.2^2}+\frac{5}{2^2.3^2}+\frac{7}{3^2.4^2}+...+\frac{199}{99^2.100^2}\). So sánh \(M\) với \(1\)
CM:
\(\dfrac{3}{1^2.2^2}+\dfrac{5}{2^2.3^2}+\dfrac{7}{3^2.4^2}+...+\dfrac{19}{9^2+10^2}< 1\)
1) Tính:
a) \(\frac{\left(1+\frac{17}{1}\right).\left(1+\frac{17}{2}\right).\left(1+\frac{17}{3}\right).....\left(1+\frac{17}{19}\right)}{\left(1+\frac{19}{1}\right).\left(1+\frac{19}{2}\right).\left(1+\frac{19}{3}\right).....\left(1+\frac{19}{17}\right)}\)
b) \(\frac{\frac{-6}{5}+\frac{6}{19}-\frac{6}{23}}{\frac{9}{5}-\frac{9}{19}+\frac{9}{23}}\)
c) \(\frac{\frac{1}{6}-\frac{1}{39}+\frac{1}{51}}{\frac{1}{8}-\frac{1}{52}+\frac{1}{68}}\)
d) \(\frac{\frac{2}{3}-\frac{2}{5}-\frac{2}{7}+\frac{2}{11}}{\frac{13}{3}-\frac{13}{5}-\frac{13}{7}+\frac{13}{11}}\)
e) \(\frac{\frac{1}{1009}+\frac{1}{1010}+\frac{1}{1011}+...+\frac{1}{2017}}{1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{2015}-\frac{1}{2016}+\frac{1}{2017}}\)
2) CMR: \(\frac{1}{2^2}+\frac{1}{2^4}+\frac{1}{2^6}+...+\frac{1}{2^{800}}< \frac{1}{3}\)
Tính : \(A=\frac{2,5-4\left(\frac{5}{2}-1,2\right)+\frac{3}{8}}{4\left(\frac{5}{2}-1,2\right)-\frac{3}{5}:\frac{2}{5}}-\frac{55}{148}\)
Rút gọn \(B=\frac{3^9-2^3.3^7+2^{10}.3^2-2^{13}}{3^{10}-2^2.3^7+2^{10}.3^3-2^{12}}\)
A = \(\frac{\frac{1}{11}-\frac{1}{13}-\frac{1}{17}}{\frac{5}{11}-\frac{5}{13}-\frac{5}{17}}+\frac{\frac{2}{3}-\frac{2}{9}-\frac{2}{27}+\frac{2}{81}}{\frac{7}{3}-\frac{7}{9}-\frac{7}{27}+\frac{7}{81}}\)
B = \(\frac{5^2}{11.16}+\frac{5^2}{16.21}+\frac{5^2}{21.26}+\frac{5^2}{26.31}+...+\frac{5^2}{56.81}\)
C = \(1-\frac{1}{3}-\frac{1}{6}-\frac{1}{10}-\frac{1}{15}-...-\frac{1}{1225}\)