Question

I. Tìm x,biết :

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

( \(x\in N,x\ge2)\)

Answer 1

*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 \(x\) = \(12\) .

Answer 2

Answer 3

I.

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

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

\(\Leftrightarrow\frac{1}{2}.\left(\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{6}+...+\frac{1}{2x-2}-\frac{1}{2x}\right)=\frac{11}{48}\)

\(\Leftrightarrow\frac{1}{2}-\frac{1}{2x}=\frac{11}{24}\)

\(\Leftrightarrow\frac{1}{2x}=\frac{1}{2}-\frac{11}{24}\)

\(\Leftrightarrow\frac{1}{2x}=\frac{1}{24}\)

\(\Leftrightarrow2x=24\)

\(\Leftrightarrow x=24:2\)

\(\Rightarrow x=12\left(TM\right)\)

Vậy \(x=12.\)

Chúc bạn học tốt!