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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, \(\Leftrightarrow\orbr{\begin{cases}2x^2-13=-5\\2x^2-13=5\end{cases}\Leftrightarrow\orbr{\begin{cases}2x^2=8\\2x^2=18\end{cases}\Leftrightarrow}\orbr{\begin{cases}x^2=4\\x^2=9\end{cases}\Leftrightarrow}\orbr{\begin{cases}x=\pm2\\x=\pm3\end{cases}}}\)
\(a)=\frac{7}{25}+\frac{4}{13}-\frac{5}{2}+\frac{18}{25}-\frac{17}{13}\)
\(=1-1-\frac{5}{2}\)
\(=-\frac{5}{2}\)
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: \(100^{99}+1>100^{68}+1\Rightarrow\frac{1}{100^{99}+1}< \frac{1}{100^{68}+1}\Rightarrow\frac{-99}{100^{99}+1}>\frac{-99}{100^{68}+1}\)
\(\Rightarrow100+\frac{-99}{100^{99}+1}>100+\frac{-99}{100^{68}+1}\Rightarrow\frac{100^{100}+1}{100^{99}+1}>\frac{100^{69}+1}{100^{68}+1}\)
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
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