\(\left(2024-x\right)^2=1-y^2\)

=>\(\left(2024-x\right)^2+y^2=1\)

mà x,y nguyên

nên \(\left(2024-x\right)^2+y^2=0+1=1+0\)

TH1: \(\left\{{}\begin{matrix}\left(2024-x\right)^2=0\\y^2=1\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}2024-x=0\\y\in\left\{1;-1\right\}\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}x=2024\\y\in\left\{1;-1\right\}\end{matrix}\right.\)

TH2: \(\left\{{}\begin{matrix}\left(2024-x\right)^2=1\\y^2=0\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}2024-x\in\left\{1;-1\right\}\\y=0\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}x\in\left\{2023;2025\right\}\\y=0\end{matrix}\right.\)

\(15\%+1,1:\left(\dfrac{2}{5}-1\dfrac{1}{2}\right)-\left(-\dfrac{1}{3}\right)^2\)

\(=\dfrac{3}{2}-\dfrac{1}{9}+1,1:\left(0,4-1,5\right)\)

\(=\dfrac{27-2}{18}+1,1:\left(-1,1\right)\)

\(=\dfrac{25}{18}-1=\dfrac{7}{18}\)

1/3 . 6/(-7) = -6/21 = -2/7

\(\dfrac{1}{3}\). \(\dfrac{6}{-7}\) = \(\dfrac{ }{7}\)

   \(\dfrac{2}{-7}\) =  \(\dfrac{ }{7}\)

        \(◻\) = \(\dfrac{2}{-7}\) x 7

       \(◻\)  = \(-2\) 

-2/3 . -5/8 = 10/24 = 5/12

\(\dfrac{-2}{3}\).\(\dfrac{-5}{8}\) = \(\dfrac{ }{12}\)

\(\dfrac{5}{12}\) = \(\dfrac{◻}{12}\)

\(◻\) = \(\dfrac{5}{12}\) \(\times\) 12

\(◻\)  = 5

\(\dfrac{-2}{3}.\dfrac{-5}{8}\) = \(\dfrac{5}{12}\)

d; \(\dfrac{x}{468}\) = \(\dfrac{-7}{13}\).\(\dfrac{5}{9}\) 

     \(\dfrac{x}{468}\) = \(\dfrac{-35}{117}\)

      \(x\)    = \(\dfrac{-35}{117}\) \(\times\) 468

      \(x\)   = - 140

 Vậy \(x=-140\)

e; \(\dfrac{2}{3}.x\) - \(\dfrac{4}{7}=\dfrac{1}{8}\)

    \(\dfrac{2}{3}.x\)         =  \(\dfrac{1}{8}\) + \(\dfrac{4}{7}\)

     \(\dfrac{2}{3}\).\(x\)        = \(\dfrac{39}{56}\)

        \(x\)         = \(\dfrac{39}{56}\) : \(\dfrac{2}{3}\)

        \(x\)          = \(\dfrac{117}{112}\)

Vậy \(x\) = \(\dfrac{117}{112}\) 

f; \(\dfrac{-2}{3}\) : (\(\dfrac{1}{2}\) - 3\(x\)) = \(\dfrac{5}{3}\)

             \(\dfrac{1}{2}\) - 3\(x\)  = \(\dfrac{-2}{3}\) : \(\dfrac{5}{3}\)

             \(\dfrac{1}{2}\) - 3\(x\) = \(\dfrac{-2}{5}\)

                   3\(x\)  = \(\dfrac{1}{2}\) + \(\dfrac{2}{5}\)

                   3\(x\)  = \(\dfrac{9}{10}\)

                      \(x\) = \(\dfrac{9}{10}\) : 3

                      \(x\)  = \(\dfrac{3}{10}\)

Vậy \(x=\dfrac{3}{10}\)

            Giải:

   n ⋮ 9 ⇔ 7 + a + 5 + 8 + b + 4  ⋮ 9

             (7 + 5 + 8 + 4) + (a + b) ⋮ 9

                      24 + (a + b) ⋮ 9

                       a + b - 3 ⋮ 9 (1)

                       a - b  = 6

                       a  = 6 + b

              Thay a  = 6 + b vào biểu thức  (1)

               6 + b  + b - 3 ⋮ 9

                      2b + 3 ⋮ 9

                ⇒ 2b + 3 \(\in\) B(9) = {0; 9; 18; 27; 36;..;}

Lập bảng ta có:  

Theo bảng trên ta có:  b  = 3; Thay b = 3 vào biểu thức a = 6 + b

ta có: a  = 6 + 3  = 9 

Vậy (a; b) = (9; 3)

A = \(\dfrac{1}{2.3}\) + \(\dfrac{1}{3.4}\) + \(\dfrac{1}{4.5}\) + \(\dfrac{1}{5.6}\) + \(\dfrac{1}{6.7}\)

A = \(\dfrac{1}{2}\) - \(\dfrac{1}{3}\) + \(\dfrac{1}{3}\) - \(\dfrac{1}{4}\) + \(\dfrac{1}{4}\) - \(\dfrac{1}{5}\) + \(\dfrac{1}{5}\) - \(\dfrac{1}{6}\) + \(\dfrac{1}{6}\) - \(\dfrac{1}{7}\)

A = \(\dfrac{1}{2}\) - \(\dfrac{1}{7}\)

A = \(\dfrac{5}{14}\)

1/(2.3) + 1/(3.4) + 1/(4.5) + 1/(5.6) + 1/(6.7)

= 1/2 - 1/3 + 1/3 - 1/4 + 1/4 - 1/5 + 1/5 + 1/6 + 1/6 - 1/7

= 1/2 - 1/7

= 5/14

Đặt \(A=\dfrac{1}{2}+\dfrac{1}{2^2}+...+\dfrac{1}{2^9}\)

=>\(2A=1+\dfrac{1}{2}+...+\dfrac{1}{2^8}\)

=>\(2A-A=1+\dfrac{1}{2}+...+\dfrac{1}{2^8}-\dfrac{1}{2}-\dfrac{1}{2^2}-...-\dfrac{1}{2^9}\)

=>\(A=1-\dfrac{1}{2^9}=\dfrac{2^9-1}{2^9}=\dfrac{511}{512}\)