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Ta có:
\(\frac{3}{5.2!}+\frac{3}{5.3!}+\frac{3}{5.4!}+...+\frac{3}{5.100!}\)
\(=\frac{3}{5}.\left(\frac{1}{2!}+\frac{1}{3!}+\frac{1}{4!}+...+\frac{1}{100!}\right)\)
\(< \frac{3}{5}.\left(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{99.100}\right)\)
\(=\frac{3}{5}.\left(1-\frac{1}{100}\right)\)
\(< \frac{3}{5}.1=\frac{3}{5}=0,6\)
\(=\frac{5.2^{30}.3^{18}-2^2.3^{20}.2^{27}}{5.2^9.2^{19}.3^{19}+7.2^{29}.3^{18}}\)
\(=\frac{5.2^{30}.3^{18}-2^{29}.3^{20}}{5.2^{28}.3^{19}+7.2^{29}.3^{18}}\)
\(=\frac{2^{29}.3^{18}\left(5.2-3^2\right)}{2^{28}.3^{18}\left(5.3+7.2\right)}\)
\(=\frac{2}{29}\)
a.ta có :4^2.4^3/2^10=2^4.2^6/2^10=2^10/2^10=1
b. ta co :(0,6)^5/(0,2)^5=(0,6/0,2)^5=3^5=243
a) Ta có: \(\frac{0,4-\frac{2}{7}+\frac{2}{11}}{0,6-\frac{3}{7}+\frac{3}{11}}\)
\(=\frac{2.\left(0.2-\frac{1}{7}+\frac{1}{11}\right)}{3.\left(0,2-\frac{1}{7}+\frac{1}{11}\right)}\)
\(=\frac{2}{3}\)
b) Ta có: \(\frac{0,75-0,6+\frac{3}{7}+\frac{3}{13}}{2,75-2,2+\frac{11}{7}+\frac{11}{13}}\)
\(=\frac{3.\left(0,25-0,2+\frac{1}{7}+\frac{1}{13}\right)}{11.\left(0,25-0,2+\frac{1}{7}+\frac{1}{13}\right)}\)
\(=\frac{3}{11}\)
Chuk pạn hok tốt!
a) \(\frac{0,4-\frac{2}{7}+\frac{2}{11}}{0,6-\frac{3}{7}+\frac{3}{11}}\\ =\frac{\frac{2}{5}-\frac{2}{7}+\frac{2}{11}}{\frac{3}{5}-\frac{3}{7}+\frac{3}{11}}\\ =\frac{2\left(\frac{1}{5}-\frac{1}{7}+\frac{1}{11}\right)}{3\left(\frac{1}{5}-\frac{1}{7}+\frac{1}{11}\right)}\\=\frac{2}{3}\)
b) \(\frac{0,75-0,6+\frac{3}{7}+\frac{3}{13}}{2,75-2,2+\frac{11}{7}+\frac{11}{13}}\\=\frac{\frac{3}{4}-\frac{3}{5}+\frac{3}{7}+\frac{3}{13}}{\frac{11}{4}-\frac{11}{5}+\frac{11}{7}+\frac{11}{13}}\\=\frac{3\left(\frac{1}{4}-\frac{1}{5}+\frac{1}{7}+\frac{1}{13}\right)}{11\left(\frac{1}{4}-\frac{1}{5}+\frac{1}{7}+\frac{1}{13}\right)}\\=\frac{3}{11}\)
Theo đầu bài ta có:
\(\frac{3}{5\cdot2!}+\frac{3}{5\cdot3!}+\frac{3}{5\cdot4!}+...+\frac{3}{5.100!}< 0,6\)
\(\Rightarrow\frac{3}{5}\cdot\frac{1}{2!}+\frac{3}{5}\cdot\frac{1}{3!}+\frac{3}{5}\cdot\frac{1}{4!}+...+\frac{3}{5}\cdot\frac{1}{100!}< \frac{3}{5}\)
\(\Rightarrow\frac{3}{5}\cdot\left(\frac{1}{2!}+\frac{1}{3!}+\frac{1}{4!}+...+\frac{1}{100!}\right)< \frac{3}{5}\)
\(\Rightarrow\frac{1}{2!}+\frac{1}{3!}+\frac{1}{4!}+...+\frac{1}{100!}< 1\)( điều cần chứng minh )
Mà \(\frac{1}{2!}+\frac{1}{3!}+\frac{1}{4!}+...+\frac{1}{100!}< \frac{1}{1\cdot2}+\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+...+\frac{1}{99\cdot100}\)
\(\Rightarrow\frac{1}{2!}+\frac{1}{3!}+\frac{1}{4!}+...+\frac{1}{100!}< 1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{99}-\frac{1}{100}\)
\(\Rightarrow\frac{1}{2!}+\frac{1}{3!}+\frac{1}{4!}+...+\frac{1}{100!}< 1-\frac{1}{100}< 1\)( đã chứng minh được )
Vậy \(\frac{3}{5\cdot2!}+\frac{3}{5\cdot3!}+\frac{3}{5\cdot4!}+...+\frac{3}{5\cdot100!}< 0,6\)( đpcm )