a, Vì \(\left(x-2\right)^2\ge0\) nên \(A=\left(x-2\right)^2+24\ge24\)

Dấu '=' xảy ra khi và chỉ khi: \(\left(x-2\right)^2=0\Leftrightarrow x=2\)

Vậy GTNN của A là 24 khi x=2.

b,Vì \(-x^2\le0\) nên \(B=-x^2+\dfrac{13}{5}\le\dfrac{13}{5}\)

Dấu '=' xảy ra khi và chỉ khi: \(-x^2=0\Leftrightarrow x=0\)

Vậy GTLN của B là \(\dfrac{13}{5}\) khi x=0

Ai trả lời nhanh và đúng mik give tick xanh nhé.

Nhanh giúp mình nhé, mình đang cần gấp.

ai giúp mình với😥

B = \(\dfrac{1}{2^3}\) + \(\dfrac{2}{3^3}\) + \(\dfrac{3}{4^3}\)+...+ \(\dfrac{n-1}{n^3}\) (n > 2)

Vì n > 2 ⇒ B > 0 (1)

   \(\dfrac{1}{2^3}\) < \(\dfrac{1}{2^2}\) < \(\dfrac{1}{1.2}\) = \(\dfrac{1}{1}\) - \(\dfrac{1}{2}\)

    \(\dfrac{2}{3^3}\) < \(\dfrac{3}{3^3}\) = \(\dfrac{1}{3^2}\) < \(\dfrac{1}{2.3}\) = \(\dfrac{1}{2}\) - \(\dfrac{1}{3}\)

    \(\dfrac{3}{4^3}\)  <  \(\dfrac{4}{4^3}\) =  \(\dfrac{1}{4^2}\) < \(\dfrac{1}{3.4}\) = \(\dfrac{1}{3}\) - \(\dfrac{1}{4}\)

     ..................................................

    \(\dfrac{n-1}{n^3}\)<\(\dfrac{n^{ }}{n^3}\)  = \(\dfrac{1}{n^2}\) < \(\dfrac{1}{\left(n-1\right).n}\) = \(\dfrac{1}{n-1}\) - \(\dfrac{1}{n}\)

Cộng vế với vế ta có: 

 B <  1 - \(\dfrac{1}{n}\) < 1 (2)

Kết hợp (1) và(2) ta có: 0 < B < 1

Vậy B không phải là số tự nhiên (đpcm)

a, -4\(\dfrac{3}{5}\).2\(\dfrac{4}{3}\) < \(x\) < -2\(\dfrac{3}{5}\): 1\(\dfrac{6}{15}\)

  - \(\dfrac{23}{5}\).\(\dfrac{10}{3}\) <   \(x\)   < - \(\dfrac{13}{5}\): \(\dfrac{21}{15}\)

   -  \(\dfrac{46}{3}\)     <  \(x\) < - \(\dfrac{13}{7}\) 

          \(x\) \(\in\) {-15; -14;-13;..; -2}

a) Ta có \(-4\dfrac{3}{5}\cdot2\dfrac{4}{3}=-\dfrac{23}{5}\cdot\dfrac{10}{3}=-\dfrac{46}{3}\) và \(-2\dfrac{3}{5}\div1\dfrac{6}{15}=-\dfrac{13}{5}\div\dfrac{7}{5}=-\dfrac{13}{7}\)

Do đó \(-\dfrac{46}{3}< x< -\dfrac{13}{7}\)

Lại có \(-\dfrac{46}{3}\le-15\) và \(-\dfrac{13}{7}\ge-2\)

Suy ra \(-15\le x\le-2\), x ϵ Z

b) Ta có \(-4\dfrac{1}{3}\left(\dfrac{1}{2}-\dfrac{1}{6}\right)=-\dfrac{13}{3}\cdot\dfrac{1}{3}=-\dfrac{13}{9}\) và \(-\dfrac{2}{3}\left(\dfrac{1}{3}-\dfrac{1}{2}-\dfrac{3}{4}\right)=-\dfrac{2}{3}\cdot\dfrac{-11}{12}=\dfrac{11}{18}\)

Do đó \(-\dfrac{13}{9}< x< \dfrac{11}{18}\)

Lại có \(-\dfrac{13}{9}\le-1\) và \(\dfrac{11}{18}\ge0\)

Suy ra \(-1\le x\le0\), x ϵ Z

`#3107`

`a)`

\(\dfrac{1}{1\cdot2}+\dfrac{1}{2\cdot3}+\dfrac{1}{3\cdot4}+...+\dfrac{1}{1999\cdot2000}\)

\(=1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{1999}-\dfrac{1}{2000}\)

\(=1-\dfrac{1}{2000}\)

\(=\dfrac{1999}{2000}\)

`b)`

\(\dfrac{1}{1\cdot4}+\dfrac{1}{4\cdot7}+\dfrac{1}{7\cdot10}+...+\dfrac{1}{100\cdot103}?\)

\(=\dfrac{1}{3}\cdot\left(\dfrac{3}{1\cdot4}+\dfrac{3}{4\cdot7}+\dfrac{3}{7\cdot10}+...+\dfrac{3}{100\cdot103}\right)\)

\(=\dfrac{1}{3}\cdot\left(1-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{7}+...+\dfrac{1}{100}-\dfrac{1}{103}\right)\)

\(=\dfrac{1}{3}\cdot\left(1-\dfrac{1}{103}\right)\)

\(=\dfrac{1}{3}\cdot\dfrac{102}{103}\)

\(=\dfrac{34}{103}\)

`c)`

\(\dfrac{8}{9}-\dfrac{1}{72}-\dfrac{1}{56}-\dfrac{1}{42}-....-\dfrac{1}{6}-\dfrac{1}{2}\)

\(=\dfrac{8}{9}-\left(\dfrac{1}{2}+\dfrac{1}{6}+...+\dfrac{1}{42}+\dfrac{1}{56}+\dfrac{1}{72}\right)\)

\(=\dfrac{8}{9}-\left(\dfrac{1}{1\cdot2}+\dfrac{1}{2\cdot3}+...+\dfrac{1}{6\cdot7}+\dfrac{1}{7\cdot8}+\dfrac{1}{8\cdot9}\right)\)

\(=\dfrac{8}{9}-\left(1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{8}-\dfrac{1}{9}\right)\)

\(=\dfrac{8}{9}-\left(1-\dfrac{1}{9}\right)\)

\(=\dfrac{8}{9}-\dfrac{8}{9}\\ =0\)

b) Sửa đề:

 \(\dfrac{1}{1.4}+\dfrac{1}{4.7}+\dfrac{1}{7.10}+...+\dfrac{1}{100.103}\)

\(=\dfrac{1}{3}.\left(1-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{10}+...+\dfrac{1}{100}-\dfrac{1}{103}\right)\)

\(=\dfrac{1}{3}.\left(1-\dfrac{1}{103}\right)\)

\(=\dfrac{1}{3}.\left(\dfrac{103}{103}-\dfrac{1}{103}\right)\)

\(=\dfrac{1}{3}.\dfrac{102}{103}\)

\(=\dfrac{34}{103}\)

\(-\dfrac{3}{4}.31\dfrac{11}{23}-0,75.8\dfrac{12}{23}\)

\(=-\dfrac{3}{4}.31\dfrac{11}{23}-\dfrac{3}{4}.8\dfrac{12}{23}\)

\(=-\dfrac{3}{4}.\left(31+\dfrac{11}{23}+8+\dfrac{12}{12}\right)\)

\(=-\dfrac{3}{4}.\left(31+8+1\right)\)

\(=-\dfrac{3}{4}.40\)

\(=-3.10\)

\(=-30\)

- \(\dfrac{3}{4}\).31\(\dfrac{11}{23}\) - 0,75.8\(\dfrac{12}{23}\)

= - \(\dfrac{3}{4}\).\(\dfrac{724}{23}\) - \(\dfrac{3}{4}\)\(\dfrac{196}{23}\)

=   - \(\dfrac{3}{4.23}.\left(724+196\right)\)

= - \(\dfrac{3}{92}\) . 920

= - 30 

`(3/4)^5 * x=(3/4)^7`

`=> x= (3/4)^7 : (3/4)^5`

`=> x= (3/4)^(7-5)`

`=>x=(3/4)^2`

`=>x= 9/16`

\(\left(\dfrac{3}{4}\right)^5\cdot x=\left(\dfrac{3}{4}\right)^7\)

\(x=\left(\dfrac{3}{4}\right)^7\div\left(\dfrac{3}{4}\right)^5=\left(\dfrac{3}{4}\right)^2\)

Gọi diện tích mỗi thửa ruộng theo thứ tự từ bé đến lớn lần lượt là: \(x\); y (đk : \(x\); y > 0)

Theo bài ra ta có: \(\dfrac{x}{3.3}\) = \(\dfrac{y}{4.4}\)  = \(\dfrac{x}{9}\) = \(\dfrac{y}{16}\) 

Áp dụng tính chất dãy tỉ số bằng nhau ta có: 

                             \(\dfrac{x}{9}\)  = \(\dfrac{y}{16}\) = \(\dfrac{x+y}{9+16}\) = \(\dfrac{10000}{25}\) = 400

⇒ \(x\) = 400 x 9 = 3 600 (m²)

    y = 400 x 16 = 6 400 (m²)

Kết luận .......