`#3107.101107`

\(\dfrac{5}{7}\times\dfrac{6}{11}+\dfrac{5}{11}\times\dfrac{1}{7}-\dfrac{5}{7}\times\dfrac{14}{11}\\ =\dfrac{5}{7}\times\dfrac{6}{11}+\dfrac{5}{7}\times\dfrac{1}{11}-\dfrac{5}{7}\times\dfrac{14}{11}\\ =\dfrac{5}{7}\times\left(\dfrac{6}{11}+\dfrac{1}{11}-\dfrac{14}{11}\right)\\ =\dfrac{5}{7}\times\left(-\dfrac{7}{11}\right)\\ =-\dfrac{5}{11}\)

\(\dfrac{5}{7}\cdot\dfrac{6}{11}+\dfrac{5}{11}\cdot\dfrac{1}{7}-\dfrac{5}{7}\cdot\dfrac{14}{11}\)

\(=\dfrac{5}{7}\cdot\dfrac{6}{11}+\dfrac{5}{7}\cdot\dfrac{1}{11}-\dfrac{5}{7}\cdot\dfrac{14}{11}\)

\(=\dfrac{5}{7}\cdot\left(\dfrac{6}{11}+\dfrac{1}{11}-\dfrac{14}{11}\right)\)

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

3\(^{x+1}\) - 3\(^x\) = 1428

3\(^x\).( 3 - 1) = 1428

3\(^x\).2        = 1428

3\(^x\)           = 1428: 2

3\(^x\)           = 714 

3\(^{x+1}\) là số lẻ \(\forall\) \(x\) ⇒ 3\(^x\) \(\ne\) 714 ∀ \(x\) ⇒ \(x\) \(\in\) \(\varnothing\)

`#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

?

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)