*Bài làm:

~I) Tìm x:

➤Ta có: \(\frac{1}{2.4}\) + \(\frac{1}{4.6}\) + ... + \(\frac{1}{\left(2x-2\right)2x}\) = \(\frac{11}{48}\)

⇒ \(2\) . (\(\frac{1}{2.4}\) + \(\frac{1}{4.6}\) + ... + \(\frac{1}{\left(2x-2\right)2x}\)) = \(2\) . \(\frac{11}{48}\)

⇒ \(\frac{2}{2.4}\) + \(\frac{2}{4.6}\) + ... + \(\frac{2}{\left(2x-2\right)2x}\) = \(\frac{22}{48}\)

⇒ (\(\frac{1}{2}\) - \(\frac{1}{4}\)) + (\(\frac{1}{4}\) - \(\frac{1}{6}\)) + ... + (\(\frac{1}{2x-2}\) - \(\frac{1}{2x}\)) = \(\frac{22}{48}\)

⇒ \(\frac{1}{2}\) - \(\frac{1}{4}\) + \(\frac{1}{4}\) - \(\frac{1}{6}\) + \(\frac{1}{6}\) - ... - \(\frac{1}{2x-2}\) + \(\frac{1}{2x-2}\) - \(\frac{1}{2x}\) = \(\frac{22}{48}\)

⇒ \(\frac{1}{2}\) - \(\frac{1}{2x}\) = \(\frac{22}{48}\)

⇒ \(\frac{x}{x}\) . \(\frac{1}{2}\) - \(\frac{1}{2x}\) = \(\frac{22}{48}\)

⇒ \(\frac{x}{2x}\) - \(\frac{1}{2x}\) = \(\frac{22}{48}\)

⇒ \(\frac{x-1}{2x}\) = \(\frac{22}{48}\)

⇒ \(\frac{x-1}{2x}\) = \(\frac{22}{48}\)

⇒ \(x-1\) = \(\frac{22}{48}\) . \(2x\)

⇒ \(x-1\) = \(\frac{44x}{48}\)

⇒ \(x\) = \(\frac{44x}{48}\) + \(1\)

⇒ \(x\) = \(\frac{44x}{48}\) + \(\frac{48}{48}\)

⇒ \(x\) = \(\frac{44x+48}{48}\)

⇒ \(x\) = \(12\) (Chỗ này mình bấm máy tính nên hơi tắt;Bạn thông cảm)

*Vậy \(x\) = \(12\) .

\(\frac{1}{2.4}+\frac{1}{4.6}+...+\frac{1}{\left(2x-2\right).2x}=\frac{1}{8}\)

\(\Rightarrow\frac{1}{2}\left(\frac{2}{2.4}+\frac{2}{4.6}+...+\frac{2}{\left(2x-2\right).2x}\right)=\frac{1}{8}\)

\(\Rightarrow\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{6}+...+\frac{1}{2x-2}-\frac{1}{2x}=\frac{1}{8}:\frac{1}{2}\)

\(\Rightarrow\frac{1}{2}-\frac{1}{2x}=\frac{1}{4}\)

\(\Rightarrow\frac{1}{2x}=\frac{1}{2}-\frac{1}{4}=\frac{1}{4}\)

\(\Leftrightarrow2x=4\)

\(\Leftrightarrow x=2\)

TL:
\(\frac{1}{2}\left(\frac{2}{2.4}+\frac{2}{4.6}+....+\frac{2}{\left(2x-2\right)2x}\right)=\frac{1}{8}\)  

\(\frac{1}{2}-\frac{1}{4x}=\frac{1}{8}\) 

\(\frac{1}{4x}=\frac{3}{8}\) 

=>x=2/3

hc tốt

\(\frac{1}{2.5}+\frac{1}{5.8}+...+\frac{1}{\left(3n-1\right)\left(3n+2\right)}\)

\(=\frac{1}{3}.\left(\frac{3}{2.5}+\frac{3}{5.8}+...+\frac{3}{\left(3n-1\right)\left(3n+2\right)}\right)\)

\(=\frac{1}{3}.\left(\frac{1}{2}-\frac{1}{5}+\frac{1}{5}-\frac{1}{8}+...+\frac{1}{3n-1}-\frac{1}{3n+2}\right)\)

\(=\frac{1}{3}.\left(\frac{1}{2}-\frac{1}{3n+2}\right)\)

\(=\frac{1}{3}.\frac{3n}{2.\left(3n+2\right)}\)

\(=\frac{n}{2\left(3n+2\right)}\)

\(\frac{1}{2.5}+\frac{1}{5.8}+...+\frac{1}{\left(3n-1\right).\left(3n+2\right)}=\frac{1}{3}.\left(\frac{3}{2.5}+\frac{3}{5.8}+...+\frac{3}{\left(3n-1\right).\left(3n+2\right)}\right)\)

                                                                          \(=\frac{1}{3}.\left(\frac{1}{2}-\frac{1}{5}+\frac{1}{5}-\frac{1}{8}+...+\frac{1}{3n-1}-\frac{1}{3n+2}\right)\)

                                                                            \(=\frac{1}{3}.\left(\frac{1}{2}-\frac{1}{3n+2}\right)\)

                                                                              \(=\frac{1}{3}.\left(\frac{3n+2}{2.\left(3n+2\right)}-\frac{2}{2.\left(3n+4\right)}\right)\)

                                                                                \(=\frac{1}{3}.\frac{3n}{2.\left(3n+2\right)}=\frac{n}{2.\left(3n+2\right)}\)

Đặt A=1/2.5+1/5.8+...+1/(3n-1).(3n+2)

=>3A=3/2.5+3/5.8+...+3/(3n-1).(3n+2)

=>3A=1/2-1/5+1/5-1/8+...+1/3n-1-1/3n+2

=>3A=1/2-1/3n+2

=>3A=(3n+2-2)/[2.(3n+2)]

=>3A=3n/6n+4

=>A=3n/6n+4/3

=>A=n/6n+4

210

Đặt \(A=\frac{1}{2.5}+\frac{1}{5.8}+\frac{1}{8.11}+......+\frac{1}{\left(3n-1\right)\left(3n+2\right)}\)

\(=>3A=\frac{3}{2.5}+\frac{3}{5.8}+\frac{3}{8.11}+....+\frac{3}{\left(3n-1\right)\left(3n+2\right)}\)

=> \(3A=\frac{1}{2}-\frac{1}{5}+\frac{1}{5}-\frac{1}{8}+....+\frac{1}{3n-1}-\frac{1}{3n+2}\)

=>\(3A=\frac{1}{2}-\frac{1}{3n+2}\)

=> \(3A=\frac{\left(3n+2\right):2}{3n+2}-\frac{1}{3n+2}\)

=> \(3A=\frac{1,5.n}{3n+2}\)

=>\(A=\frac{1,5.n}{3n+2}.\frac{1}{3}=>A=\frac{1,5.n}{\left(3n+2\right).3}=\frac{1,5.n}{9n+6}\)

\(Hay\) \(A=\frac{1,5n:1,5}{\left(9n+6\right):1,5}=\frac{n}{9n:1,5+6:1,5}=\frac{n}{6n + 4} \left(đpcm\right)\)

\(\frac{1}{2\cdot4}+\frac{1}{4\cdot6}+...+\frac{1}{\left(2x-2\right)\cdot2x}=\frac{1}{8}\left(x\inℕ;x\ge2\right)\)

Đặt \(A=\frac{1}{2\cdot4}+\frac{1}{4\cdot6}+...+\frac{1}{\left(2x-2\right)2x}\)

\(2A=\frac{2}{2\cdot4}+\frac{2}{4\cdot6}+...+\frac{2}{\left(2x-2\right)2x}\)

\(2A=\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{6}+....+\frac{1}{2x-2}-\frac{1}{2x}\)

\(2A=\frac{1}{2}-\frac{1}{2x}=\frac{x-1}{2x}\)

\(\Rightarrow A=\frac{x-1}{2x}:2=\frac{x-1}{2x}\cdot\frac{1}{2}=\frac{x-1}{4x}\)

Mà \(A=\frac{1}{8}\Rightarrow\frac{x-1}{4}=\frac{1}{8}\)

\(\Leftrightarrow8x-8=4\)

\(\Leftrightarrow8x=12\)

\(\Leftrightarrow x=\frac{12}{8}=\frac{3}{2}\left(ktm\right)\)

Vậy không có x thỏa mãn yêu cầu đề bài

1. D= 1/3 + 1/3.4 + 1/3.4.5 + 1/3.4.5....n < 1/2 + 1/3.4 + 1/4.5 + ...+ 1/ n.(n-1)

=> còn lại thì bạn có thể tự chứng minh

mk chả hiểu j