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

Để A nhỏ nhất thì (x + 3)² + 1 nhỏ nhất

Ta có:

(x + 3)² ≥ 0

⇒ (x + 3)² + 1 ≥ 1

⇒ A nhỏ nhất là -5/1 = -5 khi x = -3

\(A=\dfrac{-5}{\left(x+3\right)^2+1}\) (Tìm số nguyên \(x\) để \(A_{min}\))

Vì \(\left(x+3\right)^2\ge0\forall x\)

\(\Rightarrow\left(x+3\right)^2+1\ge1\forall x\) 

\(\Rightarrow\dfrac{-5}{\left(x+3\right)^2+1}\ge-5\forall x\)

hay \(A\ge-5\)

Dấu \("="\) xảy ra:

\(\Leftrightarrow\left(x+3\right)^2=0\)

\(\Leftrightarrow x+3=0\)

\(\Leftrightarrow x=0-3=-3\left(TM\right)\)

Vậy \(M\in A=-5\Leftrightarrow x=-3\)

(\(x+2\)).(\(x^2\) + 1) ≥ 0

\(x^2\) ≥ 0 ∀ \(x\)

\(x^2\) + 1  ≥ 1 ∀ \(x\)

Lập bảng ta có:

Theo bảng trên ta có: 

\(x\) ≥ -2

Vậy \(x\) ≥ -2 

giúp mình với

chứng minh nhỏ hơn 1 nhé mọi người 

em viết thiếu

Ta có:

(a + 4b) ⋮ 13

⇒ 9(a + 4b) ⋮ 13

⇒ (9a + 36b) ⋮ 13

⇒ (9a + 36b + a + 4b) ⋮ 13

⇒ (10a + 40b) ⋮ 13

Lại có: 39b ⋮ 13

⇒ (10a + 40b - 39b) ⋮ 13

⇒ (10a + b) ⋮ 13

Mà (a + 4b) ⋮ 13

⇒ (a + 4b)(10a + b) ⋮ 13.13

⇒ (a + 4b)(10a + b) ⋮ 169

\(a+4b⋮13\Rightarrow11.\left(a+4b\right)=11a+44b⋮13\)

\(\Rightarrow\left(11a+44b\right)-\left(a+4b\right)=10a+40b=\left(10a+b\right)+39⋮13\)

Mà \(39⋮13\Rightarrow10a+b⋮13\)

Đặt 

\(a+4b=13p;10a+b=13q\)

\(\Rightarrow\left(a+4b\right).\left(10a+b\right)=13p.13q=169pq⋮169\)

Gọi x là số cần tìm (x ∈ ℕ* và 100 < x < 1000)

Do khi chia x cho 25; 28; 35 thì được các số dư lần lượt là 4; 7; 14 nên x + 21 chia hết cho 25; 28; 35

⇒ x + 21 ∈ BC(25; 28; 35)

Ta có:

25 = 5²

28 = 2².7

35 = 5.7

⇒ BCNN(25; 28; 35) = 2².5².7 = 700

⇒ x + 21 ∈ BC(25; 28; 35) = B(700) = (0; 700; 1400; ...)

⇒ x ∈ {-21; 679; 1379; ...}

Mà 100 < x < 1000

⇒ x = 679

Vậy số tự nhiên cần tìm là 679

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

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

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

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

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

12A =  3-1+\(\dfrac{1}{3}\)-\(\dfrac{1}{3^2}\)+....+\(\dfrac{1}{3^{97}}\)-\(\dfrac{1}{3^{98}}\)-\(\dfrac{100}{3^{99}}\)

12A+4A=3-1+\(\dfrac{1}{3}\)-\(\dfrac{1}{3^2}\)+..+\(\dfrac{1}{3^{97}}\)-\(\dfrac{1}{3^{98}}\)-\(\dfrac{100}{3^{99}}\)+1-\(\dfrac{1}{3}\)+\(\dfrac{1}{3^2}\)-\(\dfrac{1}{3^3}\)+..+\(\dfrac{1}{3^{98}}\)-\(\dfrac{1}{3^{99}}\)-\(\dfrac{100}{3^{100}}\)

16A = 3+(-1+1)+(\(\dfrac{1}{3}-\dfrac{1}{3}\))+...+(-\(\dfrac{1}{3^{98}}\)+\(\dfrac{1}{3^{98}}\))+(-\(\dfrac{100}{3^{99}}\)-\(\dfrac{1}{3^{99}}\)) - \(\dfrac{100}{3^{100}}\)

16A = 3 - \(\dfrac{101}{3^{99}}\)  - \(\dfrac{100}{3^{100}}\)

16A = 3 - \(\dfrac{303}{3^{100}}\) - \(\dfrac{100}{3^{100}}\)

16A = 3 - \(\dfrac{403}{3^{100}}\)

A = \(\dfrac{3}{16}\) - \(\dfrac{403}{16.3^{100}}\) < \(\dfrac{3}{16}\) < \(\dfrac{3}{14}\) (đpcm)

A = \(\dfrac{n+1}{n+2}\) ( n ≠ -2)

Gọi ƯCLN(n + 1; n + 2) = d

Ta có: \(\left\{{}\begin{matrix}\left(n+1\right)⋮d\\\left(n+2\right)⋮d\end{matrix}\right.\)

      ⇒ ( (n + 2) - (n + 1) ) ⋮ d

           (n + 2 - n - 1) ⋮ d

                             1 ⋮ d

Vậy ƯCLN(n +1; n + 2) = 1

Hay A = \(\dfrac{n+1}{n+2}\) là phân số tối giản.