** Bổ sung điều kiện $x,y$ là số nguyên.

Lời giải:

$2x+xy+y=5$

$\Rightarrow (2x+xy)+y=5$

$\Rightarrow x(y+2)+(y+2)=7$

$\Rightarrow (x+1)(y+2)=7$

Do $x,y$ là số nguyên nên $x+1, y+2$ cũng là số nguyên. Mà $(x+1)(y+2)=7$ nên ta có các TH sau:

TH1: $x+1=1, y+2=7$

$\Rightarrow x=0; y=5$

TH2: $x+1=-1, y+2=-7$

$\Rightarrow x=-2; y=-9$

TH3: $x+1=7, y+2=1$

$\Rightarrow x=6; y=-1$

TH4: $x+1=-7, y+2=-1$

$\Rightarrow x=-8; y=-3$

Tìm \(x;y\) \(\in\) Z/ 2\(x+xy+y=5\)

       Ta có: 2\(x+xy+y=5\)

              ⇒ \(x\)(2 + y) + y = 5

                    \(x\)(2 + y)  = 5 - y 

                     \(x\)           = \(\dfrac{5-y}{2+y}\) (y ≠ - 2)

                     \(x\in\) Z ⇔ 5 - y ⋮ 2 + y

                             7 - 2  - y ⋮ 2 + y

                           7 - (2 + y) ⋮ 2 + y

                            7 ⋮ 2 + y

                    2 + y \(\in\) Ư(7) = {-7; -1; 1; 7}

                    Lập bảng ta có:

Theo bảng trên ta có các cặp số nguyên thỏa mãn đề bài là:

(\(x;y\)) = (-2; -9); (-8; -3); (6; -1); (0; 5)

\(C=2+2^3+2^5+\dots+2^{99}+2^{101}\)

\(2^2\cdot C=2^3+2^5+2^7+\dots+2^{101}+2^{103}\)

\(4C-C=\left(2^3+2^5+2^7+\dots+2^{101}+2^{103}\right)-\left(2+2^3+2^5+\dots+2^{99}+2^{101}\right)\)

\(3C=2^{103}-2\)

\(\Rightarrow C=\dfrac{2^{103}-2}{3}\)

Câu 4:

a: O nằm trên tia đối của tia AB

=>A nằm giữa O và B

=>OB=OA+AB

=>OB=4+6=10(cm)

M là trung điểm của OA

=>\(OM=MA=\dfrac{OA}{2}\)

N là trung điểm của OB

=>\(ON=NB=\dfrac{OB}{2}\)

Vì OA<OB

nên OM<ON

=>M nằm giữa O và N

=>OM+MN=ON

=>\(MN=ON-OM=\dfrac{OB}{2}-\dfrac{OA}{2}=\dfrac{10}{2}-\dfrac{4}{2}=5-2=3\left(cm\right)\)

b: \(MN=ON-OM=\dfrac{OB}{2}-\dfrac{OA}{2}=\dfrac{1}{2}\left(OB-OA\right)\)

\(=\dfrac{1}{2}\cdot AB\) không đổi khi O di chuyển trên tia đối của tia AB

b;    1 + \(\dfrac{1}{3}\) + \(\dfrac{1}{6}\) + \(\dfrac{1}{10}\) + ... + \(\dfrac{2}{x\left(x+1\right)}\) = 1\(\dfrac{2023}{2025}\)

     \(\dfrac{1}{2}\).(1 + \(\dfrac{1}{3}\) + \(\dfrac{1}{6}\) + \(\dfrac{1}{10}\) + ... + \(\dfrac{2}{x\left(x+1\right)}\)) = \(\dfrac{4048}{2025}\).\(\dfrac{1}{2}\)

     \(\dfrac{1}{1.2}\) + \(\dfrac{1}{2.3}\) + \(\dfrac{1}{12}\) + \(\dfrac{1}{20}\) + ... + \(\dfrac{1}{x\left(x+1\right)}\) = \(\dfrac{2024}{2025}\)

      \(\dfrac{1}{1.2}\) + \(\dfrac{1}{2.3}\) + \(\dfrac{1}{3.4}\) + \(\dfrac{1}{4.5}\) + ... + \(\dfrac{1}{x.\left(x+1\right)}\) = \(\dfrac{2024}{2025}\)

       \(\dfrac{1}{1}\) - \(\dfrac{1}{2}\) + \(\dfrac{1}{2}\) - \(\dfrac{1}{3}\) + \(\dfrac{1}{3}\) - \(\dfrac{1}{4}\) + \(\dfrac{1}{4}\) - \(\dfrac{1}{5}\) + ... + \(\dfrac{1}{x}\) - \(\dfrac{1}{x+1}\) = \(\dfrac{2024}{2025}\)

        \(\dfrac{1}{1}\) - \(\dfrac{1}{x+1}\) = \(\dfrac{2024}{2025}\)

                \(\dfrac{1}{x+1}\) = 1  - \(\dfrac{2024}{2025}\)

                   \(\dfrac{1}{x+1}\) = \(\dfrac{1}{2025}\)

                    \(x+1\) = 2025

                     \(x\) = 2025 - 1

                      \(x=2024\)

Vậy   \(x=2024\)

\(S=\dfrac{2}{3}\times\dfrac{4}{5}\times...\times\dfrac{4046}{4047}\)

\(S< \dfrac{3}{4}\times\dfrac{5}{6}\times...\times\dfrac{4047}{4048}\)

\(S^2< \dfrac{2}{3}\times\dfrac{4}{5}\times...\times\dfrac{4046}{4047}\times\left(\dfrac{3}{4}\times\dfrac{5}{6}\times...\times\dfrac{4047}{4048}\right)\)

\(S^2< \dfrac{2\times3\times4\times5\times...\times4046\times4047}{3\times4\times5\times6\times...\times4047\times4048}\)

\(S^2< \dfrac{2}{4048}\)

⇒ \(S^2< \dfrac{1}{2024}\)

\(C=\dfrac{4}{3\cdot5}+\dfrac{4}{5\cdot7}+\dots+\dfrac{4}{97\cdot99}\)

\(=2\cdot\left(\dfrac{2}{3\cdot5}+\dfrac{2}{5\cdot7}+\dots+\dfrac{2}{97\cdot99}\right)\)

\(=2\cdot\left(\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+\dots+\dfrac{1}{97}-\dfrac{1}{99}\right)\)

\(=2\cdot\left(\dfrac{1}{3}-\dfrac{1}{99}\right)\)

\(=2\cdot\dfrac{32}{99}=\dfrac{64}{99}\)

\(D=\dfrac{18}{2\cdot5}+\dfrac{18}{5\cdot8}+\dots+\dfrac{18}{203\cdot206}\)

\(=6\cdot\left(\dfrac{3}{2\cdot5}+\dfrac{3}{5\cdot8}+\dots+\dfrac{3}{203\cdot206}\right)\)

\(=6\cdot\left(\dfrac{1}{2}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{8}+\dots+\dfrac{1}{203}-\dfrac{1}{206}\right)\)

\(=6\cdot\left(\dfrac{1}{2}-\dfrac{1}{206}\right)\)

\(=6\cdot\dfrac{51}{103}=\dfrac{306}{103}\)

Khi đó: \(\dfrac{C}{D}=\dfrac{\dfrac{64}{99}}{\dfrac{306}{103}}=\dfrac{3296}{15147}\)

=1/1-1/1+1/2-1/2+......

=0

\(10A=\dfrac{10^{11}+10}{10^{11}+1}=1+\dfrac{9}{10^{11}+1}\)

\(10B=\dfrac{10^{10}+10}{10^{10}+1}=1+\dfrac{9}{10^{10}+1}\)

\(10^{11}+1>10^{10}+1\)

=>\(\dfrac{9}{10^{11}+1}< \dfrac{9}{10^{10}+1}\)

=>\(\dfrac{9}{10^{11}+1}+1< \dfrac{9}{10^{10}+1}+1\)

=>10A<10B

=>A<B

A = \(\dfrac{10^{10}+1}{10^{11}+1}\) < \(\dfrac{10^{10}+1+9}{10^{11}+1+9}\) = \(\dfrac{10^{10}+10}{10^{11}+10}\) = \(\dfrac{10.\left(10^9+1\right)}{10.\left(10^{10}+1\right)}\) = B

Vậy A < B

\(M=\dfrac{1}{1\cdot5}+\dfrac{1}{5\cdot9}+...+\dfrac{1}{n\left(n+4\right)}\)

\(=\dfrac{1}{4}\left(\dfrac{4}{1\cdot5}+\dfrac{4}{5\cdot9}+...+\dfrac{4}{n\left(n+4\right)}\right)\)

\(=\dfrac{1}{4}\left(1-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{9}+...+\dfrac{1}{n}-\dfrac{1}{n+4}\right)\)

\(=\dfrac{1}{4}\left(1-\dfrac{1}{n+4}\right)=\dfrac{1}{4}\cdot\dfrac{n+4-1}{n+4}=\dfrac{n+3}{4\left(n+4\right)}\)

\(\dfrac{2}{3}\) + \(x+\dfrac{1}{4}\)\(x\) = - \(\dfrac{22}{27}\)

       \(x+\dfrac{1}{4}x\) = - \(\dfrac{22}{27}\) - \(\dfrac{2}{3}\)

        \(\dfrac{5}{4}x\)      = - \(\dfrac{40}{27}\)

            \(x=-\dfrac{40}{27}:\dfrac{5}{4}\)

            \(x=-\dfrac{32}{27}\)

Vậy \(x=-\dfrac{32}{27}\)