giải nè

được chưa cho mình k

Ta có:

\(A=\dfrac{1}{1.1981}+\dfrac{1}{2.1982}+...+\dfrac{1}{n\left(1980+n\right)}+...+\dfrac{1}{25.2005}\)

\(=\dfrac{1}{1980}\left(\dfrac{1981-1}{1.1981}+\dfrac{1982-2}{2.1982}+...+\dfrac{1980+n-n}{n\left(1980+n\right)}+...+\dfrac{2005-25}{25.2005}\right)\)

\(=\dfrac{1}{1980}\left(1-\dfrac{1}{1981}+\dfrac{1}{2}-\dfrac{1}{1982}+...+\dfrac{1}{n}-\dfrac{1}{1980+n}+...+\dfrac{1}{25}-\dfrac{1}{2005}\right)\)

\(=\dfrac{1}{1980}\left[\left(1+\dfrac{1}{2}+...+\dfrac{1}{25}\right)-\left(\dfrac{1}{1981}+\dfrac{1}{1982}+...+\dfrac{1}{2005}\right)\right]\)

Lại có:

\(B=\dfrac{1}{1.26}+\dfrac{1}{2.27}+...+\dfrac{1}{m\left(m+25\right)}+...+\dfrac{1}{1980.2005}\)

\(=\dfrac{1}{25}\left(\dfrac{26-1}{1.26}+\dfrac{27-2}{2.27}+...+\dfrac{25+m-m}{m\left(25+m\right)}+...+\dfrac{2005-1980}{1980.2005}\right)\)

\(=\dfrac{1}{25}\left(\dfrac{1}{1}-\dfrac{1}{26}+\dfrac{1}{2}-\dfrac{1}{27}+...+\dfrac{1}{m}-\dfrac{1}{25+m}+...+\dfrac{1}{1980}-\dfrac{1}{2005}\right)\)

\(=\dfrac{1}{25}\left[\left(\dfrac{1}{1}+\dfrac{1}{2}+...+\dfrac{1}{1980}\right)-\left(\dfrac{1}{26}+\dfrac{1}{27}+...+\dfrac{1}{2005}\right)\right]\)

\(=\dfrac{1}{25}\left[\left(1+\dfrac{1}{2}+...+\dfrac{1}{25}\right)-\left(\dfrac{1}{1981}+\dfrac{1}{1982}+...+\dfrac{1}{2005}\right)\right]\)

\(\Rightarrow\dfrac{A}{B}=\dfrac{\dfrac{1}{1980}}{\dfrac{1}{25}}=\dfrac{5}{396}\)

Vậy tỉ số của \(A\) và \(B\) là \(\dfrac{5}{396}\)

1, \(=\frac{3\left(\frac{1}{3}-\frac{1}{5}+\frac{1}{11}+\frac{1}{13}\right)}{7\left(\frac{1}{3}-\frac{1}{5}+\frac{1}{11}+\frac{1}{13}\right)}=\frac{3}{7}\)

2, a, \(\Leftrightarrow\left(3x-2\right)^{10}-\left(3x-2\right)^6=0\)

\(\Leftrightarrow\left(3x-2\right)^6\left[\left(3x-2\right)^4-1\right]=0\)

TH1: (3x-2)^6=0 <=> 3x-2=0 <=> x=2/3

TH2: (3x-2)^4-1=0 <=> (3x-2)^4=1

<=> 3x-2 = 1 hoặc 3x-2=-1

<=>x=1 hoặc x=-1/3

Vậy x=2/3 hoặc x=1 hoặc x=-1/3

b, \(\Leftrightarrow\orbr{\begin{cases}2x^2-13=-5\\2x^2-13=5\end{cases}\Leftrightarrow\orbr{\begin{cases}2x^2=8\\2x^2=18\end{cases}\Leftrightarrow}\orbr{\begin{cases}x^2=4\\x^2=9\end{cases}\Leftrightarrow}\orbr{\begin{cases}x=\pm2\\x=\pm3\end{cases}}}\)

3,a, \(A=\frac{3n+9}{n-4}=\frac{3n-12+21}{n-4}=\frac{3\left(n-4\right)+21}{n-4}=3+\frac{21}{n-4}\)

Để \(A\in Z\Leftrightarrow\frac{21}{n-4}\in Z\Leftrightarrow21⋮n-4\Leftrightarrow n-4\inƯ\left(21\right)\)

Ta có bảng

Vậy..

b, tương tự a

\(a)=\frac{7}{25}+\frac{4}{13}-\frac{5}{2}+\frac{18}{25}-\frac{17}{13}\)

\(=1-1-\frac{5}{2}\)

\(=-\frac{5}{2}\)

cái này bạn bấm máy tính là ra mà 

Bài 1: \(\left(\frac{-1}{16}\right)^{100}=\frac{1}{\left(2^4\right)^{100}}=\frac{1}{2^{400}}>\frac{1}{2^{500}}=\left(\frac{-1}{2}\right)^{500}.\)

Bài 2: \(100^{99}+1>100^{68}+1\Rightarrow\frac{1}{100^{99}+1}< \frac{1}{100^{68}+1}\Rightarrow\frac{-99}{100^{99}+1}>\frac{-99}{100^{68}+1}\)

\(\Rightarrow100+\frac{-99}{100^{99}+1}>100+\frac{-99}{100^{68}+1}\Rightarrow\frac{100^{100}+1}{100^{99}+1}>\frac{100^{69}+1}{100^{68}+1}\)

$\textbf{a)}$

$\left(-\dfrac34+\dfrac27\right):\dfrac27+\left(-\dfrac14+\dfrac57\right):\dfrac23$

$=\left(-\dfrac{13}{28}\right)\cdot\dfrac72+\dfrac{13}{28}\cdot\dfrac32$

$=-\dfrac{13}{8}+\dfrac{39}{56}$

$=-\dfrac{13}{14}.$

$\textbf{b)}$

$\left(-\dfrac13\right)^2\cdot\dfrac4{11}+\dfrac7{11}\cdot\left(-\dfrac13\right)^2$

$=\dfrac19\left(\dfrac4{11}+\dfrac7{11}\right)$

$=\dfrac19.$

a) 27^n : 3^n = 9

     (27 : 3)^n  = 9

         9^n        = 9

=> n = 1

b) 25/5^n = 5

       5^n     = 25 : 5

       5^n     = 5

=> n = 1

c) 81/(-3)^n = -243

         (-3)^n   = -243 : 81

         (-3)^n   = -3

=> n = 1

d) 1/2 . 2^n + 4 . 2^n = 9 . 2^5

       2^n . (1/2 + 4)     = 9 . 32

       2^n . 9/2              = 288

       2^n                       = 288 : 9/2

       2^n                       = 64

       2^n                       = 2^6

=> n = 6

thank you for doing your homework well !