Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
Ta có :\(\left(2011+1\right)^2=2011^2+1+2.2011\)
\(\Rightarrow2011^2+1=2012-2.2011\)
\(\Rightarrow N=\sqrt{2012^2-2.2011+\left(\dfrac{2011}{2012}\right)^2}+\dfrac{2011}{2012}\)
\(=\sqrt{\left(2012-\dfrac{2011}{2012}\right)^2}+\dfrac{2011}{2012}\)
\(=2012-\dfrac{2011}{2012}+\dfrac{2011}{2012}\)
\(=2019\)
Vậy N có giá trị là một số tự nhiên.
\(\dfrac{1}{1.2.3}+\dfrac{1}{2.3.4}+...+\dfrac{1}{n\left(n+1\right)\left(n+2\right)}=\dfrac{637}{2550}\)
\(\Leftrightarrow\dfrac{1}{2}\left(\dfrac{1}{1.2}-\dfrac{1}{2.3}+\dfrac{1}{2.3}-\dfrac{1}{3.4}+...+\dfrac{1}{n\left(n+1\right)}-\dfrac{1}{\left(n+1\right)\left(n+2\right)}\right)=\dfrac{637}{2550}\)
\(\Leftrightarrow\dfrac{1}{2}\left(\dfrac{1}{1.2}-\dfrac{1}{\left(n+1\right)\left(n+2\right)}\right)=\dfrac{637}{2550}\)
\(\Leftrightarrow\dfrac{1}{1.2}-\dfrac{1}{\left(n+1\right)\left(n+2\right)}=\dfrac{637}{1275}\)
\(\Leftrightarrow\dfrac{1}{\left(n+1\right)\left(n+2\right)}=\dfrac{1}{2}-\dfrac{637}{1275}=\dfrac{1}{2550}\)
\(\Leftrightarrow\left(n+1\right)\left(n+2\right)=2550\)
\(\Leftrightarrow n^2+3n-2548=0\)
\(\Rightarrow n=49\)
a: \(=\dfrac{1}{x-y}\cdot x^2\cdot\left(x-y\right)=x^2\)
b: \(=\sqrt{27\cdot48}\cdot\left|a-2\right|=36\left(a-2\right)\)
c: \(=\left(\sqrt{2012}+\sqrt{2011}\right)^2\)
d: \(=\dfrac{8}{7}\cdot\dfrac{-x}{y+1}\)
e: \(=\dfrac{11}{12}\cdot\dfrac{x}{-y-2}=\dfrac{-11x}{12\left(y+2\right)}\)
Lời giải:
\(A=\frac{1}{1.2}+\frac{2}{1.2.3}+\frac{3}{1.2.3.4}+...+\frac{2011}{1.2.3...2012}\)
\(=\frac{2-1}{1.2}+\frac{3-1}{1.2.3}+\frac{4-1}{1.2.3.4}+...+\frac{2012-1}{1.2.3...2012}\)
\(=1-\frac{1}{1.2}+\frac{1}{1.2}-\frac{1}{1.2.3}+\frac{1}{1.2.3}-\frac{1}{1.2.3.4}+...+\frac{1}{1.2.3...2011}-\frac{1}{1.2.3...2012}\)
\(=1-\frac{1}{1.2...2012}< 1\)
Ta có đpcm.