1.2.3.4.5.6.7.8.9 - 1.2.3.4.5.6.7.8 - 1.2.3.4.5.6.7 - 8²

= 1.2.3.4.5.6.7.(8.9 - 8 - 1) - 64

= 5040.63 - 64

= 317520 - 64

= 317456

\(1\times2\times3\times4\times5\times6\times7\times8\times9-1\times2\times3\times4\times5\times6\times7\times8-1\times2\times3\times4\times5\times6\times7-8^2\)

\(=1\times2\times3\times4\times5\times6\times7\times\left(8\times9-8-1\right)-64\)

\(=5040\times63-64\)

\(=317520-64\)

\(=317456\)

a) \(x=\dfrac{7}{25}+\dfrac{-1}{5}\)

\(\Rightarrow x=\dfrac{7}{25}+\dfrac{-5}{25}\)

\(\Rightarrow x=\dfrac{2}{25}\)c

b) \(x=\dfrac{5}{11}+\dfrac{4}{-9}\)

\(\Rightarrow x=\dfrac{45}{99}+\dfrac{-44}{99}\)

\(\Rightarrow x=\dfrac{1}{99}\)

c) \(x-\dfrac{5}{7}=\dfrac{1}{9}\)

\(\Rightarrow x=\dfrac{1}{9}+\dfrac{5}{7}\)

\(\Rightarrow x=\dfrac{7}{63}+\dfrac{45}{63}\)

\(\Rightarrow x=\dfrac{52}{63}\) 

d) \(x-\dfrac{5}{7}=\dfrac{2}{7}\cdot\dfrac{11}{5}\)

\(\Rightarrow x-\dfrac{5}{7}=\dfrac{22}{35}\)

\(\Rightarrow x=\dfrac{22}{35}+\dfrac{5}{7}\)

\(\Rightarrow x=\dfrac{22}{35}+\dfrac{25}{35}\)

\(\Rightarrow x=\dfrac{47}{35}\)

e) \(\dfrac{3}{4}-x=1\)

\(\Rightarrow x=\dfrac{3}{4}-1\)

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

f) x + 4 = 1/5

x = 1/5 - 4

x = 1/5 - 20/5

x = -19/5

g) x - 1/5 = 2

x = 2 + 1/5

x = 10/5 + 1/5

x = 11/5

h) x + 5/3 = 1/27

x = 1/27 - 5/3

x = 1/27 - 45/27

x = -44/27

i) x/15 = 3/5 + (-2/3)

x/15 = 9/15 - 10/15

x/15 = -1/15

x = -1

k) 3x/4 = 18/(3x + 1)

3x(3x + 1) = 18.4 (1)

Đặt t = 3x

(1) ⇒ t(t + 1) = 72

t² + t - 72 = 0

t² - 8t + 9t - 72 = 0

(t² - 8t) + (9t - 72) = 0

t(t - 8) + 9(t - 8) = 0

(t - 8)(t + 9) = 0

t - 8 = 0 hoặc t + 9 = 0

*) t - 8 = 0

t = 8

3x = 8

x = 8/3

*) t + 9 = 0

t = -9

3x = -9

x = -9 : 3

x = -3

Vậy x = -3; x = 8/3

a; \(x\) = \(\dfrac{7}{25}\) + \(\dfrac{-1}{5}\)

     \(x\) = \(\dfrac{7}{25}\) - \(\dfrac{5}{25}\)

      \(x=\dfrac{2}{25}\)

b; \(x=\dfrac{5}{11}\) + \(\dfrac{4}{-9}\)

    \(x=\dfrac{45}{99}-\dfrac{44}{99}\)

     \(x=\dfrac{1}{99}\)

c; \(x\) - \(\dfrac{5}{7}\) = \(\dfrac{1}{9}\)

   \(x\)        = \(\dfrac{1}{9}\) + \(\dfrac{5}{7}\)

    \(x=\dfrac{7}{63}+\dfrac{45}{63}\) 

   \(x\) = \(\dfrac{52}{63}\)

a; \(\dfrac{x}{7}\) = \(\dfrac{9}{y}\) (\(x>y\))

    \(x.y\) = 7.9

    \(xy\)  = 63

Ư(63) = {-63;-21 -9; 7; -3; -1; 1; 3; 7; 9;21; 63}

Lập bảng ta có:

Vì \(x>y\) nên theo bảng trên ta có các cặp số nguyên \(x;y\) thỏa mãn đề bài là:

(\(x;y\)) = (-7; -9); (-3; -21); (-1; -63); (9; 7); (21; 3); (63; 1)

b; \(\dfrac{x}{15}\) = \(\dfrac{3}{y}\) Và \(x< y< 0\)

    \(x.y\) = 3.15 

    \(xy\)   = 45

45 = 3².5;  Ư(45) = {-45; -15; -9; -5; -3; -1; 1; 3; 5; 9; 15; 45}

Lập bảng ta có:

Vì \(x< y< 0\)

Theo bảng trên ta có:

các cặp \(x;y\) nguyên thỏa mãn đề bài là:

(-45; -1); (-15; -3); (-9; -5)

Bài 3.17 

a = \(\dfrac{n+8}{2n-5}\)  (n \(\in\) N*)

a \(\in\) Z ⇔ n + 8  ⋮ 2n - 5

           2.(n + 8) ⋮ 2n - 5

            2n + 16 ⋮ 2n - 5

       2n - 5 + 21 ⋮ 2n - 5

                     21 ⋮ 2n - 5

        2n - 5 \(\in\) Ư(21)

21 = 3.7; Ư(21) = {-21; -7; -3; -1; 1; 3; 7; 21}

Lập bảng ta có: 

Theo bảng trên ta có: n = 6

Vậy n = 6 thì a là số nguyên tố.

\(\dfrac{7n-1}{4}\) \(\in\) N ; \(\dfrac{5n+3}{12}\) \(\in\) N 

⇔ \(\left\{{}\begin{matrix}7n-1⋮4\\5n+3⋮12\end{matrix}\right.\)

⇒ \(\left\{{}\begin{matrix}3.\left(7n-1\right)⋮12\\5n+3⋮12\end{matrix}\right.\)

⇒  \(\left\{{}\begin{matrix}21n-3⋮12\\5n+3⋮12\end{matrix}\right.\)

⇒  21n - 3 + 5n + 3 ⋮ 12

      (21n + 5n) ⋮ 12

       26n ⋮ 12

       13n ⋮ 6

        n ⋮ 6

⇒ 7n là số chẵn ⇒ 7n -  1 là số lẻ nên  7n - 1 không chia hết cho 4 

Vậy không tồn tại số tự nhiên n nào thỏa mãn đề bài.

\(S=\dfrac{1}{1!}+\dfrac{1}{2!}+....+\dfrac{1}{2001!}\)

\(S=1+\dfrac{1}{2!}+\dfrac{1}{3!}+.....+\dfrac{1}{2001!}\)

\(\dfrac{1}{2!}=\dfrac{1}{1\times2};\dfrac{1}{3!}< \dfrac{1}{2\times3};...;\dfrac{1}{2001!}< \dfrac{1}{2000\times2001}\)

\(\dfrac{1}{2!}+\dfrac{1}{3!}+....+\dfrac{1}{2001!}< \dfrac{1}{1\times2}+\dfrac{1}{2\times3}+....+\dfrac{1}{2000\times2001}\)

\(S< 1+1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{2000}-\dfrac{1}{2001}\)

\(S< 2-\dfrac{1}{2001}< 2< 3\)

=> \(S< 3\)

A = \(\dfrac{2023^{2024^{2025}}-2017^{2024^{2023}}}{10}\)

A = \(\dfrac{2023^{2^{2025}.1012^{2025}}-2017^{2^{2023}.1012^{2023}}}{10}\)

A = \(\dfrac{2023^{2^2.2^{2023}.1012^{2025}}-2017^{2^2.2^{2021}1012^{2023}.}}{10}\)

A = \(\dfrac{2023^{4.2^{2023}.1012^{2025}}-2017^{4.2^{2021}.1012^{2023}}}{10}\)

A = \(\dfrac{\left(2023^4\right)^{2^{2023}.1012^{2025}}-\left(2017^4\right)^{2^{2021}.1012^{2023}}}{10}\)

A = \(\dfrac{\left(\overline{..1}\right)^{2^{2023}.1012^{2025}}-\left(\overline{..1}\right)^{2^{2021}.1012^{2023}}}{10}\)

A = \(\dfrac{\overline{..1}-\overline{..1}}{10}\)

A = \(\dfrac{\overline{..0}}{10}\) 

A  \(\in\) N (đpcm)

A = \(\dfrac{2023^{2024^{2025}}-2017^{2024^{2023}}}{10}\)

A = \(\dfrac{2023^{2^{2025}.1012^{2025}}-2017^{2^{2023}.1012^{2023}}}{10}\)

A = \(\dfrac{2023^{2^2.2^{2023}.1012^{2025}}-2017^{2^2.2^{2021}1012^{2023}.}}{10}\)

A = \(\dfrac{2023^{4.2^{2023}.1012^{2025}}-2017^{4.2^{2021}.1012^{2023}}}{10}\)

A = \(\dfrac{\left(2023^4\right)^{2^{2023}.1012^{2025}}-\left(2017^4\right)^{2^{2021}.1012^{2023}}}{10}\)

A = \(\dfrac{\left(\overline{..1}\right)^{2^{2023}.1012^{2025}}-\left(\overline{..1}\right)^{2^{2021}.1012^{2023}}}{10}\)

A = \(\dfrac{\overline{..1}-\overline{..1}}{10}\)

A = \(\dfrac{\overline{..0}}{10}\) 

A  \(\in\) N (đpcm)

\(5x+xy-4y=3\)

\(\Rightarrow x\left(y+5\right)-4y-20=3-20\)

\(\Rightarrow x\left(y+5\right)-4\left(y+5\right)=-17\)

\(\Rightarrow\left(y+5\right)\left(x-4\right)=-17\)

Bổ sung: \(x,y\in Z\) 

Ta có bảng:

Vậy: ... 

\(5x+xy-4y=3\)

\(\Rightarrow x\cdot\left(y+5\right)-4y-20=3-20\)

\(\Rightarrow x\cdot\left(y+5\right)-4\cdot\left(y+5\right)=-17\)

\(\Rightarrow\left(y+5\right)\cdot\left(x-4\right)=-17\)

\(\Leftrightarrow x,y\in Z\)

Lập bảng giá trị:

Vậy \(\left(x;y\right)\in\left\{\left(21;-6\right),\left(-13;-4\right),\left(3;12\right),\left(5;-22\right)\right\}\)