Ta có:

\(\frac{3}{5.2!}+\frac{3}{5.3!}+\frac{3}{5.4!}+...+\frac{3}{5.100!}\)

\(=\frac{3}{5}.\left(\frac{1}{2!}+\frac{1}{3!}+\frac{1}{4!}+...+\frac{1}{100!}\right)\)

\(< \frac{3}{5}.\left(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{99.100}\right)\)

\(=\frac{3}{5}.\left(1-\frac{1}{100}\right)\)

\(< \frac{3}{5}.1=\frac{3}{5}=0,6\)

bang nhau

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

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 )

\(\frac{4^5.4}{3^5.3}.\frac{6^5.6}{2^5.2}\)

= \(\frac{4^6}{3^6}.\frac{6^6}{2^6}\)

= \(\left(\frac{4.6}{3.2}\right)^6\)

= \(4^6\)

=> \(4^6\) = \(8^{x}\)

\(2^{12}\) = \(2^{3x}\)

12= 3.x

x= 12:3

x= 4