So sánh: \(\frac{3}{5.2!}+\frac{3}{5.3!}+\frac{3}{5.4!}+...+\frac{3}{5.100!}\) với 0,6
CM
\(\frac{3}{5.2!}\)+\(\frac{3}{5.3!}\)+\(\frac{3}{5.4!}\)+ ..... +\(\frac{3}{5.100!}\)<\(0,6\)
BT1: Tinh
\(1.A=\left(4-\frac{1}{2}+\frac{2}{3}\right)+\left(5+\frac{4}{3}-\frac{6}{5}\right)-\left(6-\frac{7}{4}+\frac{4}{5}\right)\)
\(2.B=\frac{\left(-1\right)^6.3^5.4^3}{9^2.2^5}\)
\(3.\frac{4}{5}.\frac{11}{3}-\frac{4}{5}.\frac{8}{3}+\frac{1}{5}\)
\(4.\sqrt{289-\sqrt{169+\sqrt{256-\sqrt{196}}}}\)
\(5.\frac{3^{15}.2^{18}.5^4}{6^{14}.10^5}\)
Chứng minh rằng:
a. \(\frac{1}{3^2}+\frac{2}{3^3}+\frac{3}{3^4}+\frac{4}{3^5}+...+\frac{99}{3^{100}}+\frac{100}{3^{101}}< \frac{1}{4}\)
b.\(\frac{1}{2}-\frac{1}{4}+\frac{1}{8}-\frac{1}{16}+\frac{1}{32}-\frac{1}{64}< \frac{1}{3}\)
c.\(\frac{1}{3}-\frac{2}{3^2}+\frac{3}{3^3}-\frac{4}{3^4}+...+\frac{99}{3^{99}}-\frac{100}{3^{100}}< \frac{1}{16}\)
d. \(\frac{1}{5^2}-\frac{2}{5^3}+\frac{3}{5^4}-\frac{4}{5^5}+...+\frac{99}{5^{100}}-\frac{100}{5^{101}}< \frac{1}{36}\)
CMR
a)A=\(\frac{1}{3}+\frac{2}{3^2}+\frac{3}{3^3}+....+\frac{100}{3^{100}}< \frac{3}{4}\)
b)B=\(\frac{1}{3}+\frac{2}{3^2}+\frac{3}{3^3}+.....+\frac{100}{4^{100}}< \frac{4}{9}\)
CMR: \(C=\frac{1}{3}+\frac{2}{3^2}+\frac{3}{3^3}+\frac{4}{3^4}+...+\frac{100}{3^{100}}< \frac{3}{4}\)
CMR:
\(\frac{1}{3}+\frac{2}{3^2}+\frac{3}{3^3}+\frac{4}{3^4}+.....+\frac{100}{3^{100}}< \frac{3}{4}\)
so sánh :
C=\(\frac{5^4.20^4}{25^5.4^5}\) và D=\(\left(\frac{-10}{3}\right)^5.\left(\frac{-6}{5}\right)^4\)
E=\(\left(1+\frac{2}{3}-\frac{1}{4}\right).\left(\frac{4}{5}-\frac{3}{4}\right)^2\)và F=\(2:\left(\frac{1}{2}-\frac{2}{3}\right)^3\)
I=\(\frac{2}{3}+\frac{1}{3}:\left(\frac{-8}{25}\right)\)và H=\(\frac{5}{11}.\frac{4}{11}+\frac{7}{11}.\frac{5}{11}-\frac{2}{3}\)