\(\dfrac{1}{2!}+\dfrac{2}{3!}+\dfrac{3}{4!}+...+\dfrac{n}{\left(n+1\right)!}\)

\(=\dfrac{2-1}{2!}+\dfrac{3-1}{3!}+\dfrac{4-1}{4!}+...+\dfrac{\left(n+1\right)-1}{\left(n+1\right)!}\)

\(=\dfrac{2}{2!}-\dfrac{1}{2!}+\dfrac{3}{3!}-\dfrac{1}{3!}+...+\dfrac{\left(n+1\right)}{\left(n+1\right)!}-\dfrac{1}{\left(n+1\right)!}\)

\(=1-\dfrac{1}{2!}+\dfrac{1}{2!}-\dfrac{1}{3!}+\dfrac{1}{3!}+...+\dfrac{1}{n!}-\dfrac{1}{\left(n+1\right)!}\)

( Vì \(\dfrac{3}{3!}=\dfrac{1}{2!};\dfrac{4}{4!}=\dfrac{1}{3!};...;\dfrac{n+1}{\left(n+1\right)!}=\dfrac{1}{n!}\))

\(=1-\dfrac{1}{\left(n+1\right)!}< 1\)

Đặt \(S\left(n\right)=\dfrac{1}{2!}+\dfrac{2}{3!}+\dfrac{3}{4!}+...+\dfrac{n}{\left(n+1\right)!}\)

Ta có \(S\left(1\right)=\dfrac{1}{2!}=\dfrac{1}{2}=1-\dfrac{1}{2!}\)

\(S\left(2\right)=S\left(1\right)+\dfrac{2}{3!}=\dfrac{1}{2}+\dfrac{1}{3}=\dfrac{5}{6}=1-\dfrac{1}{3!}\)

\(S\left(3\right)=S\left(2\right)+\dfrac{3}{4!}=\dfrac{5}{6}+\dfrac{1}{8}=\dfrac{23}{24}=1-\dfrac{1}{4!}\)

Từ đây, ta có \(S\left(n\right)=1-\dfrac{1}{\left(n+1\right)!}\) và hiển nhiên \(S\left(n\right)< 1\) do \(\dfrac{1}{\left(n+1\right)!}>0\)

Vậy ta có đpcm

BC của 125;25;60;8;4=3000

\(\Rightarrow\dfrac{1}{125}=\dfrac{24}{3000}\)

\(\dfrac{1}{25}=\dfrac{120}{3000}\)

\(0,125=\dfrac{1}{8}=\dfrac{375}{3000}\)

\(-\dfrac{1}{60}=-\dfrac{50}{3000}\)

\(0,04=\dfrac{1}{25}=\dfrac{120}{3000}\)

\(\dfrac{1}{4}=\dfrac{750}{3000}\)

Vậy ta có: \(-\dfrac{50}{3000};\dfrac{24}{3000};\dfrac{120}{3000};\dfrac{120}{3000};\dfrac{375}{3000};\dfrac{750}{3000}\)

tức:         \(-\dfrac{1}{60};\dfrac{1}{125};0,04;\dfrac{1}{25};0,125;\dfrac{1}{4}\)

Có chỗ nào sai cho mik xin lỗi

T/c của dãy tỉ số bằng nhau ko áp dụng được với phép `.` hoặc `:` đâu e =)

Ò, giờ em mới biết, cảm ơn cj đã nhắc!

Lời giải:

$0,(15)=\frac{15}{99}=\frac{5}{33}$

Lời giải:
1.

\(\frac{16^3.8^5}{4^{12}}=\frac{(2^4)^3.(2^3)^5}{(2^2)^{12}}=\frac{2^{12}.2^{15}}{2^{24}}=\frac{2^{27}}{2^{24}}=2^3=8\)

2. 

\(\frac{5^4.9^5}{15^3.27^2}=\frac{5^4.(3^2)^5}{3^3.5^3.(3^3)^2}=\frac{5^4.3^{10}}{5^3.3^9}=5.3=15\)

3. 

\(\frac{10^5.7^3}{14^2.20^4}=\frac{2^5.5^5.7^3}{2^2.7^2.(2^2.5)^4}=\frac{2^5.5^5.7^3}{2^{10}.5^4.7^2}=\frac{5.7}{2^5}=\frac{35}{32}\)

4.

\(\frac{8^3.6^5}{2^{12}.27^2}=\frac{(2^3)^3.2^5.3^5}{2^{12}.(3^3)^2}=\frac{2^{14}.3^5}{2^{12}.3^6}=\frac{2^2}{3}=\frac{4}{3}\)

`5/12 xx (-3/4) + 7/12 xx (-3/4)`

`=(-3/4) xx (5/12 + 7/12)`

`=(-3)/4 xx ((5+7)/12)`

`=(-3)/4 xx 12/12`

`=(-3)/4 xx 1`

`=-3/4`

____________________

`1/7 xx (-3)/8 + 1/7 xx (-13)/8`

`=1/7 xx (-3/8 + (-13)/8)`

`=1/7 xx `\(\left(\dfrac{-3+\left(-13\right)}{8}\right)\)

`=1/7xx (-16)/8`

`=1/7 xx (-2)`

`=-2/7`

________________________________
`3/5 xx 13/46 - 1/10 xx 16/23`

`= 3 xx 1/2 xx 1/5 xx 13/23 - 1/10 xx 16/23`

`= 3 xx 1/10 xx 13/23 - 1/10 xx 16/23`

`=1/10 xx ( 3xx13/23-16/23 )`

`=1/10 xx (38/23 - 16/23)`

`=1/10 xx 23/23`

`=1/10 x 1`

`=1/10`

______________________________

`2/3 - 4 xx ( 1/2 + 3/4)`

`=2/3 - 4 xx ( 2/4 + 3/4)`

`=2/3 - 4 xx 5/4`

`=2/3 - 5`

`=2/3 - 15/3`

`=-13/3`