Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
Câu 1:
\(\Leftrightarrow1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{n}-\dfrac{1}{n+1}=\dfrac{2999}{3000}\)
\(\Leftrightarrow1-\dfrac{1}{n+1}=\dfrac{2999}{3000}\)
=>n+1=3000
hay n=2999
2S=\(\dfrac{2}{1.3}+\dfrac{2}{3.5}+...+\dfrac{2}{2007.2009}\)
=\(1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-...+\dfrac{1}{2007}-\dfrac{1}{2009}\)
= 1- \(\dfrac{1}{2009}\)
= \(\dfrac{2008}{2009}\)
=> S=\(\dfrac{1004}{2009}\)
\(S=\dfrac{1}{1.3}+\dfrac{1}{3.5}+\dfrac{1}{5.7}+...+\dfrac{1}{\left(2n-1\right)\left(2n+1\right)}\)
\(\Rightarrow2S=\dfrac{2}{1.3}+\dfrac{2}{3.5}+\dfrac{2}{5.7}+...+\dfrac{2}{\left(2n-1\right)\left(2n+1\right)}\)
\(\Rightarrow2S=1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+...+\dfrac{1}{2n-1}-\dfrac{1}{2n+1}\) \(\Rightarrow2S=1-\dfrac{1}{2n+1}\)
\(\Rightarrow S=\dfrac{n}{2n+1}\)
Ta có : \(\dfrac{1}{\left(2n-1\right)\left(2n+1\right)}=\dfrac{1}{2}\left(\dfrac{1}{2n-1}-\dfrac{1}{2n+1}\right)\)
ta được \(\dfrac{1}{1.3}=\dfrac{1}{2}\left(\dfrac{1}{1}-\dfrac{1}{3}\right);\dfrac{1}{3.5}=\dfrac{1}{2}\left(\dfrac{1}{3}-\dfrac{1}{5}\right);\dfrac{1}{5.7}=\dfrac{1}{2}\left(\dfrac{1}{5}-\dfrac{1}{7}\right)\)
\(\dfrac{1}{\left(2n-1\right)\left(2n+1\right)}=\dfrac{1}{2}\left(\dfrac{1}{2n-1}-\dfrac{1}{2n+1}\right)\) vậy \(S=\dfrac{1}{2}\left(1-\dfrac{1}{2n+1}\right)=\dfrac{n}{2n+1}\)
\(1-\dfrac{3}{n\left(n+2\right)}=\dfrac{n\left(n+2\right)-3}{n\left(n+2\right)}=\dfrac{\left(n-1\right)\left(n+3\right)}{n\left(n+2\right)}\)
\(\Rightarrow M=\dfrac{1.5}{2.4}.\dfrac{2.6}{3.5}.\dfrac{3.7}{4.6}...\dfrac{\left(n-1\right)\left(n+3\right)}{n\left(n+2\right)}\)
\(=\dfrac{1.2.3...\left(n-1\right)}{2.3.4...n}.\dfrac{5.6.7...\left(n+3\right)}{4.5.6...\left(n+2\right)}\)
\(=\dfrac{1}{n}.\dfrac{n+3}{4}=\dfrac{n+3}{4n}=\dfrac{1}{4}+\dfrac{3}{4n}>\dfrac{1}{4}\) (đpcm)
b.
\(B=\dfrac{1}{1.2.3}+\dfrac{1}{2.3.4}+...+\dfrac{1}{\left(n-1\right)n\left(n+1\right)}\\ =\dfrac{1}{2}\left(\dfrac{2}{1.2.3}+\dfrac{2}{2.3.4}+....+\dfrac{2}{\left(n-1\right).n.\left(n+1\right)}\right)\\ =\dfrac{1}{2}\left(\dfrac{1}{1.2}-\dfrac{1}{2.3}+\dfrac{1}{2.3}-\dfrac{1}{3.4}+...+\dfrac{1}{\left(n-1\right).n}-\dfrac{1}{n\left(n+1\right)}\right)\\ =\dfrac{1}{2}\left(\dfrac{1}{2}-\dfrac{1}{n\left(n+1\right)}\right)=\dfrac{1}{4}-\dfrac{1}{2n\left(n+1\right)}\)
a) ĐKXĐ: \(x\ne1\)
Ta có: \(x^2-8x+7=0\)
\(\Leftrightarrow x^2-x-7x+7=0\)
\(\Leftrightarrow x\left(x-1\right)-7\left(x-1\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left(x-7\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-1=0\\x-7=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\left(loại\right)\\x=7\left(nhận\right)\end{matrix}\right.\)
Thay x=7 vào B, ta được:
\(B=\dfrac{1}{7-1}=\dfrac{1}{6}\)
Vậy: Khi \(x^2-8x+7=0\) thì \(B=\dfrac{1}{6}\)
b) Ta có: \(A=\dfrac{x^2+2}{x^3-1}+\dfrac{x+1}{x^2+x+1}\)
\(=\dfrac{x^2+2}{\left(x-1\right)\left(x^2+x+1\right)}+\dfrac{\left(x+1\right)\left(x-1\right)}{\left(x^2+x+1\right)\left(x-1\right)}\)
\(=\dfrac{x^2+2+x^2-1}{x^3-1}\)
\(=\dfrac{2x^2+1}{x^3-1}\)
\(B=\dfrac{1}{1.2.3}+\dfrac{1}{2.3.4}+\dfrac{1}{3.4.5}+......+\dfrac{1}{n\left(n+1\right)\left(n+2\right)}\)
\(=\dfrac{1}{1.2}-\dfrac{1}{2.3}+\dfrac{1}{2.3}-\dfrac{1}{3.4}+....+\dfrac{1}{n\left(n+1\right)}-\dfrac{1}{\left(n+1\right)\left(n+2\right)}\)
\(=\dfrac{1}{1.2}-\dfrac{1}{\left(n+1\right)\left(n+2\right)}\)
\(=\dfrac{1}{2}-\dfrac{1}{\left(n+1\right)\left(n+2\right)}\)
Vậy..
\(B=\dfrac{1}{2}\left(\dfrac{1}{1.2}-\dfrac{1}{2.3}+\dfrac{1}{2.3}-\dfrac{1}{3.4}+...+\dfrac{1}{n\left(n+1\right)}-\dfrac{1}{\left(n+1\right)\left(n+2\right)}\right)\)
\(B=\dfrac{1}{2}\left(\dfrac{1}{1.2}-\dfrac{1}{\left(n+1\right)\left(n+2\right)}\right)=\dfrac{n^2+3n+2-2}{4\left(n+1\right)\left(n+2\right)}=\dfrac{n\left(n+3\right)}{4\left(n+1\right)\left(n+2\right)}\)
Ta có \(\dfrac{1}{2\cdot4}+\dfrac{1}{4\cdot6}+\dfrac{1}{6\cdot8}+...+\dfrac{1}{2n\left(2n+2\right)}=\dfrac{1009}{4038}\)
\(\Leftrightarrow\dfrac{2}{2\cdot4}+\dfrac{2}{4\cdot6}+\dfrac{2}{6\cdot8}+...+\dfrac{2}{2n\left(2n+2\right)}=\dfrac{1009}{2019}\)
\(\Leftrightarrow\dfrac{1}{2}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{8}+...+\dfrac{1}{2n}-\dfrac{1}{2n+2}=\dfrac{1009}{2019}\)
\(\Leftrightarrow\dfrac{1}{2}-\dfrac{1}{2n+2}=\dfrac{1009}{2019}\)
\(\Leftrightarrow\dfrac{n}{2n+2}=\dfrac{1009}{2019}\)
\(\Leftrightarrow2019n=1009\left(2n+2\right)\)
\(\Leftrightarrow2019n=2018n+2018\)
\(\Leftrightarrow n=2018\)
Bài 1a) \(\dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+...+\dfrac{1}{2018.2019}\)
\(=1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+....+\dfrac{1}{2018}-\dfrac{1}{2019}\)
\(=1-\dfrac{1}{2019}=\dfrac{2018}{2019}\)
b) \(S=\dfrac{1}{1.3}+\dfrac{1}{3.5}+\dfrac{1}{5.7}+...+\dfrac{1}{2017.2019}\)
\(2S=\dfrac{2}{1.3}+\dfrac{2}{3.5}+\dfrac{2}{5.7}+...+\dfrac{2}{2017.2019}\)
\(2S=1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+...+\dfrac{1}{2017}-\dfrac{1}{2019}\)
\(2S=1-\dfrac{1}{2019}=\dfrac{2018}{2019}\)
\(S=\dfrac{1009}{2019}\)
Còn lại bạn làm tương tự hết nhé .