`#3107`

\(\left(2x-\dfrac{1}{2}\right)^2+\dfrac{3}{7}=\dfrac{19}{8}\\ \Rightarrow\left(2x-\dfrac{1}{2}\right)^2=\dfrac{19}{8}-\dfrac{3}{7}\\ \Rightarrow\left(2x-\dfrac{1}{2}\right)^2=\dfrac{109}{56}\\ \Rightarrow\left(2x-\dfrac{1}{2}\right)^2=\left(\sqrt{\dfrac{109}{56}}\right)^2\)

\(\Rightarrow\left[{}\begin{matrix}2x-\dfrac{1}{2}=\sqrt{\dfrac{109}{56}}\\2x-\dfrac{1}{2}=-\sqrt{\dfrac{109}{56}}\end{matrix}\right.\)

Bạn xem lại đề, số lớn quá ;-;.

may ngu nhu cut y dap an la 66

?

Vì Ot là tia phân giác của \(\widehat{xOy}\) nên \(\widehat{xOt}=\widehat{tOy}=\dfrac{\widehat{xOy}}{2}=\dfrac{80^o}{2}=40^o\)

a) Số tiền nhà thầu nhận xây ngôi nhà sau khi thỏa thuận là:
360000000.(100% - 2,5%) = 351000000 (đồng)

b) Chi phí lúc đầu của khu chăn nuôi nhà thầu đưa ra là:

975000000.(100% + 2,5%) = 99937500 (đồng)

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:

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

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

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

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

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

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