\(A=1-4+4^2-4^3+...+4^{98}-4^{99}+4^{100}\)

=>\(4A=4-4^2+4^3-4^4+...+4^{99}-4^{100}+4^{101}\)

=>\(4A+A=4-4^2+4^3-...+4^{99}-4^{100}+4^{101}+1-4+4^2-...+4^{98}-4^{99}+4^{100}\)

=>\(5A=4^{101}+1\)

=>\(A=\dfrac{4^{101}+1}{5}\)

345,7

345,7 nha bạn

\(\dfrac{3}{5}\)(\(x-\dfrac{5}{4}\)) + 0,4\(x\) = - \(\dfrac{1}{2}\)

\(\dfrac{3}{5}x\) - \(\dfrac{3}{4}\) + 0,4\(x\) = - \(\dfrac{1}{2}\)

\(0,6x+0,4x\) = \(-\dfrac{1}{2}\) + \(\dfrac{3}{4}\)

     \(x\)             =  \(\dfrac{1}{4}\)

Vậy \(x=\dfrac{1}{4}\)

20%  .x+2/5=1,2 -8/5

20%. x +2/5=-2/5

20%.x =-2/5+2/5

20%.x=0

x=0/20%

x=0

Vì mik ko vt đc dấu phần trằm nên copy phần trăm trên gg mới ra mực xanh , thông cảm nha

\(1,2-20\%\cdot x+\dfrac{2}{5}=1\dfrac{3}{5}\)

=>\(1,2+0,4-0,2\cdot x=1,6\)

=>1,6-0,2x=1,6

=>0,2x=0

=>x=0

\(\dfrac{5}{6}+\dfrac{11}{12}+...+\dfrac{9701}{9702}+\dfrac{9899}{9900}\)

\(=1-\dfrac{1}{6}+1-\dfrac{1}{12}+...+1-\dfrac{1}{9702}+1-\dfrac{1}{9900}\)

\(=1-\dfrac{1}{2\cdot3}+1-\dfrac{1}{3\cdot4}+...+1-\dfrac{1}{99\cdot100}\)

\(=98-\left(\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{99}-\dfrac{1}{100}\right)\)

\(=98-\left(\dfrac{1}{2}-\dfrac{1}{10}\right)=98-\dfrac{49}{100}=97,51\)

1)\(A=\dfrac{2}{1.2}+\dfrac{2}{2.3}+\dfrac{2}{3.4}+...+\dfrac{2}{2023.2024}\)
\(=2\left(\dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+...+\dfrac{1}{2023.2024}\right)\)
\(=2\left(1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{2023}-\dfrac{1}{2024}\right)\)
\(=2\left(1-\dfrac{1}{2024}\right)\)
\(=2\cdot\dfrac{2023}{2024}=\dfrac{2023}{1012}\)
2)
a/\(\dfrac{n}{n+2}+\dfrac{5}{n+2}=\dfrac{n+5}{n+2}=\dfrac{\left(n+2\right)+3}{n+2}=1+\dfrac{3}{n+2}\)
Để \(\dfrac{n}{n+2}+\dfrac{5}{n+2}\) là số nguyên thì \(n+2\inƯ\left(3\right)=\left\{\pm1;\pm3\right\}\)
Ta có bảng sau:

Vậy để \(\dfrac{n}{n+2}+\dfrac{5}{n+2}\) nguyên thì \(n\in\left\{-5;-3;-1;1\right\}\)
b/\(S=\dfrac{1}{1.3}+\dfrac{1}{3.5}+\dfrac{1}{5.7}+...+\dfrac{1}{2023.2025}\)
\(=\dfrac{1}{2}\left(\dfrac{2}{1.3}+\dfrac{2}{3.5}+\dfrac{2}{5.7}+...+\dfrac{2}{2023.2025}\right)\)
\(=\dfrac{1}{2}\left(1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+...+\dfrac{1}{2023}-\dfrac{1}{2025}\right)\)
\(=\dfrac{1}{2}\left(1-\dfrac{1}{2025}\right)\)
\(=\dfrac{1}{2}\cdot\dfrac{2024}{2025}=\dfrac{1012}{2025}\)
 

1) \(A=\dfrac{2}{1.2}+\dfrac{2}{2.3}+...+\dfrac{2}{2023.2024}\)

   \(A=2\left(\dfrac{1}{1.2}+\dfrac{1}{2.3}+...+\dfrac{1}{2023.2024}\right)\\ A=2\left(1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{2023}-\dfrac{1}{2024}\right)\\ A=2\left(1-\dfrac{1}{2024}\right)\\ A=2\cdot\dfrac{2023}{2024}=\dfrac{2023}{1012}\)