3.(\(x\) - 2)⁴ = 45

   ( \(x\) - 2)⁴ = 45: 3

   (\(x\) - 2)⁴ = 15

    \(\left[{}\begin{matrix}x-2=\sqrt[4]{15}\\x-2=-\sqrt[4]{15}\end{matrix}\right.\)

    \(\left[{}\begin{matrix}x=2+\sqrt[4]{15}\\x=2-\sqrt[4]{15}\end{matrix}\right.\)

\(\dfrac{6}{7}< 1< \dfrac{7}{4}\Rightarrow0>-\dfrac{6}{7}>-\dfrac{7}{4}\left(1\right)\)

\(\dfrac{8}{13}=\dfrac{8.3}{13.3}=\dfrac{24}{39}< 0< \dfrac{2}{3}=\dfrac{2.13}{3.13}=\dfrac{26}{39}\left(2\right)\)

\(\left(1\right),\left(2\right)\Rightarrow-\dfrac{7}{4}< -\dfrac{6}{7}< 0< \dfrac{8}{13}< \dfrac{2}{3}\)

a, 5ⁿ⁺¹ - 5^n-1 = 125⁴.2³.3

5^n-1.(5² - 1) = 125⁴.24

5^n-1.24         = 125⁴.24

5^n-1             = 125⁴

5^n-1             = (5³)⁴

5^n-1            = 5¹²

n - 1           = 12

n                = 12 + 1

n                = 13

b,2^2n-1 + 2²ⁿ⁺² = 3.2¹¹

   2^2n-1.(1 + 2³) = 3.2¹¹

  2^2n-1.9 = 3.2¹¹

 2^2n-1      = 2¹¹: 3

2²ⁿ        = 2¹² : 3 (xem lại đề bài em nhá)

\(\dfrac{1}{3}+\dfrac{1}{6}+\dfrac{1}{10}+...+\dfrac{1}{x.\left(2x+1\right)}=\dfrac{1}{10}\)

\(\Leftrightarrow\dfrac{1}{6}+\dfrac{1}{12}+\dfrac{1}{20}+...+\dfrac{1}{2x.\left(2x+1\right)}=\dfrac{1}{20}\)

\(\Leftrightarrow\dfrac{1}{2.3}+\dfrac{1}{3.4}+\dfrac{1}{4.5}+...+\dfrac{1}{2x.\left(2x+1\right)}=\dfrac{1}{20}\)

\(\Leftrightarrow\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{5}+...+\dfrac{1}{2x}-\dfrac{1}{2x+1}=\dfrac{1}{20}\)

\(\Leftrightarrow\dfrac{1}{2}-\dfrac{1}{2x+1}=\dfrac{1}{20}\)

\(\Leftrightarrow\dfrac{1}{2x+1}=\dfrac{9}{20}\)

\(\Leftrightarrow2x+1=\dfrac{20}{9}\Leftrightarrow x=\dfrac{11}{18}\)

Em giải như XYZ olm em nhé

Sau đó em thêm vào lập luận sau:

\(x\) = \(\dfrac{11}{18}\)

Vì \(\in\) N* 

Vậy \(x\in\) \(\varnothing\)

\(\dfrac{1}{15}\) + \(\dfrac{1}{21}\) + \(\dfrac{1}{28}\) + \(\dfrac{1}{36}\) +...+ \(\dfrac{2}{x\left(x+1\right)}\) = \(\dfrac{11}{40}\) (\(x\in\) N*)

\(\dfrac{1}{2}\).(\(\dfrac{1}{15}\)+\(\dfrac{1}{21}\)+\(\dfrac{1}{28}\)+\(\dfrac{1}{36}\)+.....+ \(\dfrac{2}{x\left(x+1\right)}\)) = \(\dfrac{11}{40}\) \(\times\) \(\dfrac{1}{2}\)

\(\dfrac{1}{30}\) + \(\dfrac{1}{42}\) + \(\dfrac{1}{56}\) + \(\dfrac{1}{72}\)+...+ \(\dfrac{1}{x\left(x+1\right)}\) = \(\dfrac{11}{80}\)

\(\dfrac{1}{5.6}\) + \(\dfrac{1}{6.7}\) + \(\dfrac{1}{7.8}\)+...+ \(\dfrac{1}{x\left(x+1\right)}\) = \(\dfrac{11}{80}\)

\(\dfrac{1}{5}\) - \(\dfrac{1}{6}\) + \(\dfrac{1}{6}\) - \(\dfrac{1}{7}\) + \(\dfrac{1}{7}\) - \(\dfrac{1}{8}\) + \(\dfrac{1}{8}\)-\(\dfrac{1}{9}\)+...+ \(\dfrac{1}{x}\)-\(\dfrac{1}{x+1}\) = \(\dfrac{11}{80}\)

\(\dfrac{1}{5}\) - \(\dfrac{1}{x+1}\) = \(\dfrac{11}{80}\)

         \(\dfrac{1}{x+1}\) = \(\dfrac{1}{5}\) - \(\dfrac{11}{80}\)

           \(\dfrac{1}{x+1}\) = \(\dfrac{1}{16}\)

            \(x\) + 1 = 16

            \(x\)       = 16 - 1

             \(x\)     = 15 

Bạn xem lại đề

? tam giác ABCD

- \(\dfrac{1}{4}\) = \(-\dfrac{1.3}{4.3}\) = \(\dfrac{-3}{12}\)

- \(\dfrac{1}{5}\) = \(\dfrac{-1.3}{3.5}\) = \(\dfrac{-3}{15}\)

Ba số hữu tỉ nằm giữa hai số hữu tỉ - \(\dfrac{1}{4}\); - \(\dfrac{1}{5}\) là ba số hữu tỉ nằm giữa hai số hữu tỉ: - \(\dfrac{3}{12}\) và - \(\dfrac{3}{15}\) 

Đó lần lượt là các số hữu tỉ sau: 

-\(\dfrac{3}{13};\) - \(\dfrac{3}{14}\); 

Hữu tỉ âm hay dương bạn?

- Nếu là số hữu tỉ dương :

\(m+3>0;m-2>0\Rightarrow m>-3;m>2\Rightarrow m>2\)

- Nếu là số hữu tỉ âm :

\(m+3< 0;m-2< 0\Rightarrow m< -3;m< 2\Rightarrow m< -3\)

Để 2 số hữu tỉ đều là dương :

\(\dfrac{m+2}{5}>0\Rightarrow m>-2\left(1\right)\)

\(\dfrac{m-5}{-6}>0\Rightarrow\dfrac{5-m}{6}>0\Rightarrow m< 5\left(2\right)\)

\(\left(1\right),\left(2\right)\Rightarrow-2< m< 5\Rightarrow m\in\left\{-1;0;1;2;3;4\right\}\left(m\in Z\right)\)