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1. 3A = 3^2 + 3^3 + 3^4 + ... + 3^100 + 3^ 101
=> 3A - A = (3^2 + 3^3 + 3^4 + ... + 3^100 + 3^ 101) - (3 + 3^2 + 3^3 + 3^4 + ... + 3^100 )
=> 2A = 3^101 - 3 => 2A + 3 = 3^101 vậy n = 101
2. 2A = 8 + 2 ^ 3 + 2^4 + ... + 2^20 + 2^21
=> 2A - A = (8 + 2 ^ 3 + 2^4 + ... + 2^20 + 2^21) - (4+ 2^2 + 2 ^ 3 + 2^4 + ... + 2^20 )
=> A = 2^21 là một lũy thừa của 2
3.
a) 3A = 3 + 3^2 + 3^3 + 3^4 + ... + 3^100 + 3^ 101
=> 3A - A = (3 + 3^2 + 3^3 + 3^4 + ... + 3^100 + 3^ 101) - (1 + 3 + 3 ^2 + 3 ^ 3 + ... + 3 ^100)
=> 2A = 3^101 - 1 => A = (3^101 - 1)/2
b) 4B = 4 + 4 ^ 2 + 4 ^3 + 4 ^ 4 + ... + 4 ^ 100 + 4^ 101
=> 4B - B = (4 + 4 ^ 2 + 4 ^3 + 4 ^ 4 + ... + 4 ^ 100 + 4^ 101) - (1 + 4 + 4 ^ 2 + 4 ^3 + 4 ^ 4 + ... + 4 ^ 100 )
=> 3B = 4^101 - 1 => B = ( 4^101 - 1)/2
c) xem lại đề ý c xem quy luật như thế nào nhé.
d) 3D = 3^101 + 3^ 102 + 3^ 103 + ... + 36 150 + 3^ 151
=> 3D - D = (3^101 + 3^ 102 + 3^ 103 + ... + 36 150 + 3^ 151) - (3 ^100 + 3 ^ 101 + 3 ^ 102 + .... + 3 ^ 150)
=> 2D = 3^ 151 - 3^100 => D = ( 3^ 151 - 3^100)/2
1. 3A = 3^2 + 3^3 + 3^4 + ... + 3^100 + 3^ 101
=> 3A - A = (3^2 + 3^3 + 3^4 + ... + 3^100 + 3^ 101) - (3 + 3^2 + 3^3 + 3^4 + ... + 3^100 )
=> 2A = 3^101 - 3 => 2A + 3 = 3^101 vậy n = 101
2. 2A = 8 + 2 ^ 3 + 2^4 + ... + 2^20 + 2^21
=> 2A - A = (8 + 2 ^ 3 + 2^4 + ... + 2^20 + 2^21) - (4+ 2^2 + 2 ^ 3 + 2^4 + ... + 2^20 )
=> A = 2^21 là một lũy thừa của 2
3.
a) 3A = 3 + 3^2 + 3^3 + 3^4 + ... + 3^100 + 3^ 101
=> 3A - A = (3 + 3^2 + 3^3 + 3^4 + ... + 3^100 + 3^ 101) - (1 + 3 + 3 ^2 + 3 ^ 3 + ... + 3 ^100)
=> 2A = 3^101 - 1 => A = (3^101 - 1)/2
b) 4B = 4 + 4 ^ 2 + 4 ^3 + 4 ^ 4 + ... + 4 ^ 100 + 4^ 101
=> 4B - B = (4 + 4 ^ 2 + 4 ^3 + 4 ^ 4 + ... + 4 ^ 100 + 4^ 101) - (1 + 4 + 4 ^ 2 + 4 ^3 + 4 ^ 4 + ... + 4 ^ 100 )
=> 3B = 4^101 - 1 => B = ( 4^101 - 1)/2
c) xem lại đề ý c xem quy luật như thế nào nhé.
d) 3D = 3^101 + 3^ 102 + 3^ 103 + ... + 36 150 + 3^ 151
=> 3D - D = (3^101 + 3^ 102 + 3^ 103 + ... + 36 150 + 3^ 151) - (3 ^100 + 3 ^ 101 + 3 ^ 102 + .... + 3 ^ 150)
=> 2D = 3^ 151 - 3^100 => D = ( 3^ 151 - 3^100)/2
\(B=3^2+3^3+...+3^{99}\)
\(3B=3^3+3^4+...+3^{100}\)
\(3B-B=\left(3^3+3^4+...+3^{100}\right)-\left(3^2+3^3+...+3^{99}\right)\)
\(2B=3^{100}-3^2\)
\(B=\frac{3^{100}-9}{2}\)
\(2B+9=3^{2n+4}\)
\(\Leftrightarrow3^{2n+4}=3^{100}\)
\(\Leftrightarrow2n+4=100\)
\(\Leftrightarrow n=48\).
â) Ta có : \(2n-1⋮n+1\Leftrightarrow2n+2-2-1⋮n+1\)
\(\Leftrightarrow2\left(n+1\right)-2-1⋮n+1\)\(\Leftrightarrow2\left(n+1\right)-3⋮n+1\)
\(\Leftrightarrow2n-1⋮n+1\)khi \(3⋮n+1\Rightarrow n+1\in\)Ước của \(3\) \
\(\Leftrightarrow n+1\in\left(1;-1;3;-3\right)\)
\(\Leftrightarrow n\in\left(0;-2;2;-4\right)\)
Vậy \(n\in\left(-4;-2;0;2\right)\)
b) Ta có :\(9n+5⋮3n-2\Rightarrow3\left(3n-2\right)+6+5⋮3n-2\)
\(\Rightarrow3\left(3n-2\right)+11⋮3n-2\)
\(\Rightarrow9n+5⋮3n-2\)Khi \(11⋮3n-2\)
\(\Rightarrow3n-2\in U\left(11\right)\)
\(\Rightarrow3n-2\in\left(-11;-1;1;11\right)\)
\(\Rightarrow n\in\left(-3;1;\right)\)
Phần c) bạn tự làm nhé!
Bài 2:
a: \(\dfrac{7}{8}+x=\dfrac{3}{5}\)
=>\(x=\dfrac{3}{5}-\dfrac{7}{8}=\dfrac{24-35}{40}=\dfrac{-11}{40}\)
b: \(\dfrac{17}{2}:x=5\)
=>\(x=\dfrac{17}{2}:5\)
=>\(x=\dfrac{17}{2\cdot5}=\dfrac{17}{10}\)
c: \(x-\dfrac{3}{8}=2+\dfrac{1}{4}\)
=>\(x-\dfrac{3}{8}=\dfrac{9}{4}\)
=>\(x=\dfrac{9}{4}+\dfrac{3}{8}=\dfrac{18}{8}+\dfrac{3}{8}=\dfrac{21}{8}\)
d: \(\dfrac{1}{2}+\dfrac{3}{5}\left(x-2\right)=\dfrac{1}{5}\)
=>\(\dfrac{3}{5}\left(x-2\right)=\dfrac{1}{5}-\dfrac{1}{2}=\dfrac{-3}{10}\)
=>\(x-2=-\dfrac{1}{2}\)
=>\(x=2-\dfrac{1}{2}=\dfrac{3}{2}\)
Bài 1:
a: \(\dfrac{-3}{4}+\dfrac{1}{5}=\dfrac{-15}{20}+\dfrac{4}{20}=\dfrac{-15+4}{20}=\dfrac{-11}{20}\)
b: \(\dfrac{-2}{5}-\dfrac{1}{3}=\dfrac{-6}{15}-\dfrac{5}{15}=\dfrac{-6-5}{15}=\dfrac{-11}{15}\)
c: \(\dfrac{3}{7}\cdot\dfrac{2}{5}-\dfrac{2}{5}=\dfrac{2}{5}\left(\dfrac{3}{7}-1\right)=\dfrac{2}{5}\cdot\dfrac{-4}{7}=\dfrac{-8}{35}\)
d: \(\dfrac{1}{4}+\dfrac{3}{4}\left(\dfrac{2}{3}-\dfrac{1}{2}\right)\)
\(=\dfrac{1}{4}+\dfrac{3}{4}\cdot\dfrac{4-3}{6}\)
\(=\dfrac{1}{4}+\dfrac{3}{4}\cdot\dfrac{1}{6}=\dfrac{1}{4}+\dfrac{1}{8}=\dfrac{3}{8}\)
e: \(\dfrac{7}{2}\cdot\dfrac{8}{13}+\dfrac{8}{13}\cdot\dfrac{-5}{12}+\dfrac{8}{13}\)
\(=\dfrac{8}{13}\left(\dfrac{7}{2}-\dfrac{5}{2}+1\right)\)
\(=\dfrac{8}{13}\cdot2=\dfrac{16}{13}\)
f: \(1+\dfrac{1}{8}+\dfrac{1}{24}+\dfrac{1}{48}+\dfrac{1}{80}+\dfrac{1}{120}\)
\(=1+\dfrac{1}{2\cdot4}+\dfrac{1}{4\cdot6}+\dfrac{1}{6\cdot8}+\dfrac{1}{8\cdot10}+\dfrac{1}{10\cdot12}\)
\(=1+\dfrac{1}{2}\left(\dfrac{2}{2\cdot4}+\dfrac{2}{4\cdot6}+...+\dfrac{2}{10\cdot12}\right)\)
\(=1+\dfrac{1}{2}\left(\dfrac{1}{2}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{6}+...+\dfrac{1}{10}-\dfrac{1}{12}\right)\)
\(=1+\dfrac{1}{2}\left(\dfrac{1}{2}-\dfrac{1}{12}\right)\)
\(=1+\dfrac{1}{2}\cdot\dfrac{5}{12}=1+\dfrac{5}{24}=\dfrac{29}{24}\)