1) C/tỏ rằng :
a) \(\frac{3}{5.2!}+\frac{3}{5.3!}+\frac{3}{5.4!}+...+\frac{3}{5.100!}\) < 0,6
b) \(\frac{3}{4!}+\frac{3}{5!}+\frac{3}{6!}+...+\frac{3}{100!}\) < \(\frac{1}{3!}\)
\(\frac{3}{5.2!}+\frac{3}{5.3!}+\frac{3}{5.4!}+....+\frac{3}{5.100!}< 0,6\)
Chứng tỏ rằng:
a)\(\frac{3}{5.2!}+\frac{3}{5.3!}+\frac{3}{5.4!}+...+\frac{3}{5.100!}< \frac{3}{5}\)
b) \(\frac{1}{2!}+\frac{1}{3!}+\frac{1}{4!}+..+\frac{1}{100!}< 1\)
P/S: dấu ! nghĩa là dấu dư thừa. Vd: n! = 1x2x3x.......x n
So sánh: \(\frac{3}{5.2!}+\frac{3}{5.3!}+\frac{3}{5.4!}+...+\frac{3}{5.100!}\) với 0,6
Ta có:
\(\frac{3}{5.2!}+\frac{3}{5.3!}+\frac{3}{5.4!}+...+\frac{3}{5.100!}\)
\(=\frac{3}{5}.\left(\frac{1}{2!}+\frac{1}{3!}+\frac{1}{4!}+...+\frac{1}{100!}\right)\)
\(< \frac{3}{5}.\left(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{99.100}\right)\)
\(=\frac{3}{5}.\left(1-\frac{1}{100}\right)\)
\(< \frac{3}{5}.1=\frac{3}{5}=0,6\)
Ta có:
3/5.2!+3/5.3!+.......+3/5.100!
=3/5(1/2!+1/3!+.......+1/100!)
< 3/5(1/1.2+1/2.3+........+1/99.100)
=3/5.(1-1/100)
<3/5=0.6
=> tổng trên<0,6
CMR:
\(\frac{3}{5.2!}\)+\(\frac{3}{5.3!}\)+\(\frac{3}{5.4!}\)+ ..... +\(\frac{3}{5.100!}\)<0,6\(\frac{3}{4!}\)+\(\frac{3}{5!}\)+\(\frac{3}{6!}\)+ ..... +\(\frac{3}{100!}\)<\(\frac{1}{3!}\)CM
\(\frac{3}{5.2!}\)+\(\frac{3}{5.3!}\)+\(\frac{3}{5.4!}\)+ ..... +\(\frac{3}{5.100!}\)<\(0,6\)
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 )
\(\frac{3}{5.2!}\)+ \(\frac{3}{5.3!}\)+ \(\frac{3}{5.4!}\)+....................+ \(\frac{3}{5.100!}\)
Cho A=\(\frac{3}{5.2!}+\frac{3}{5.3!}+...+\frac{3}{5.100!}\)
C/m A<0.6
có A= \(\frac{3}{5.2!}\)+\(\frac{3}{5.3!}\)+...+\(\frac{3}{5.100!}\)=\(\frac{3}{5}\)(\(\frac{1}{2!}\)+\(\frac{1}{3!}\)+....+\(\frac{1}{100!}\))
đặt vế trong ngoặc là B. Đặt \(\frac{1}{2!}\)+\(\frac{2}{3!}\)+...+\(\frac{99}{100!}\)=C ta có C=\(\frac{2-1}{2!}\)+\(\frac{3-1}{3!}\)+....+\(\frac{100-1}{100!}\)
=\(\frac{2}{2!}\)-\(\frac{1}{2!}\)+\(\frac{1}{2!}\)-\(\frac{1}{3!}\)+...+\(\frac{1}{99!}\)-\(\frac{1}{100!}\)=1-\(\frac{1}{100!}\)<1
mà \(\frac{1}{2!}\)=\(\frac{1}{2!}\);\(\frac{1}{3!}\)<\(\frac{2}{3!}\);....;\(\frac{1}{100!}\)<\(\frac{99}{100!}\)\(\Rightarrow\)B<C<1\(\Rightarrow\)B.\(\frac{3}{5}\)<1.\(\frac{3}{5}\)=\(\frac{3}{5}\)=0.6\(\Rightarrow\)A<0.6
Cũng đơn giản mà em nhớ k cho chị nha !
S=\(\frac{3}{5.2!}+\frac{3}{5.3!}+...+\frac{3}{5.100!}\) có là số nguyên hay không vì sao
a, \(\frac{5.4^{15}.9^9-4.3^{20}.8^9}{5.2^9.6^{19}-7.2^{29}.27^6}\)
b,\(\frac{0,125-\frac{1}{5}+\frac{1}{7}}{0,375-\frac{3}{5}+\frac{3}{7}}+\frac{\frac{1}{2}+\frac{1}{3}-0,2}{\frac{3}{4}+0,5-\frac{3}{10}}\)
c,\(\frac{2}{1.4}+\frac{2}{4.7}+\frac{2}{7.10}+....+\frac{2}{97.100}\)
a) \(\frac{5.4^{15}.9^9-4.3^{20}.8^9}{5.2^9.6^{19}-7.2^{29}.27^6}\)
\(=\frac{5.2^{30}.3^{18}-2^2.2^{27}.3^{20}}{5.2^9.2^{19}.3^{19}-7.2^{29}.3^{18}}\)
\(=\frac{2^{29}.3^{18}\left(5.2-3^2\right)}{2^{18}.3^{18}\left(5.3-7.2\right)}\)
\(=\frac{2.1}{1}=2\)
b) \(\frac{0,125-\frac{1}{5}+\frac{1}{7}}{0,375-\frac{3}{5}+\frac{3}{7}}+\frac{\frac{1}{2}+\frac{1}{3}-0,2}{\frac{3}{4}+0,5}-\frac{3}{10}\)
\(=\frac{0,125-\frac{1}{5}+\frac{1}{7}}{3\left(0,125-\frac{1}{5}+\frac{1}{7}\right)}+\frac{\frac{1}{2}+\frac{1}{3}-\frac{1}{5}}{\frac{3}{4}+\frac{1}{2}-\frac{3}{10}}\)
\(=\frac{1}{3}+\frac{\frac{30}{60}+\frac{20}{60}-\frac{12}{60}}{\frac{45}{60}+\frac{30}{60}-\frac{9}{60}}\)
\(=\frac{1}{3}+\frac{\frac{19}{30}}{\frac{11}{10}}\)
\(=\frac{1}{3}+\frac{19}{33}=\frac{11}{33}+\frac{19}{33}\)
\(=\frac{30}{33}=\frac{10}{11}\)