Các câu hỏi tương tự
CMR: \(\forall n\in N\)thì \(\left|\left\{\frac{n}{1}\right\}-\left\{\frac{n}{2}\right\}+\left\{\frac{n}{3}\right\}-...-\left(-1\right)^n\left\{\frac{n}{n}\right\}\right|< \sqrt{2n}\)
Giải hệ phương trình
\(\hept{\begin{cases}16\left(\frac{1}{m}+\frac{1}{n}\right)=5\left(m+\frac{m}{n}\right)\\27\left(\frac{1}{m}+\frac{1}{m}\right)=5\left(n+\frac{n}{m}\right)\end{cases}}\)
CMR:
M=\(\frac{1}{3.\left(\sqrt{1}+\sqrt{2}\right)}\)+\(\frac{1}{5.\left(\sqrt{2}+\sqrt{3}\right)}\) +...+\(\frac{1}{\left(2n+1\right).\left(\sqrt{n}+\sqrt{n+1}\right)}< \frac{1}{2}\)
m,n >0
CMR: \(|\frac{m}{n}-\sqrt{2}|< \frac{1}{n^2\left(\sqrt{3}+\sqrt{2}\right)}\left(1\right)\)
20 tháng 6 2018 lúc 11:51
Cho \(\hept{\begin{cases}a_1>a_2>...>a_n>0\\1\le k\in Z\end{cases}}\)
CMR : \(a_1+\frac{1}{a_n\left(a_1-a_2\right)^k\left(a_2-a_3\right)^k...\left(a_{n-1}-a_n\right)^k}\ge\frac{\left(n-1\right)k+2}{\sqrt[\left(n-1\right)k+2]{k^{\left(n-1\right)k}}}\)
CMR với n thuộc N; n>=2 ta có:
\(A=\left(1-\frac{2}{6}\right)\left(1-\frac{2}{12}\right)\left(1-\frac{2}{20}\right)...\left(1-\frac{2}{n\left(n+1\right)}\right)>\frac{1}{3}\)\(\frac{1}{3}\)
a, Cm công thức
\(\forall n\ge1\) ta có \(\frac{2}{\left(2n+1\right)\left(\sqrt{n}-\sqrt{n+1}\right)}< \frac{1}{\sqrt{n}}-\frac{1}{\sqrt{n+1}}\)
b, áp dụng tính
\(\frac{1}{3\left(1+\sqrt{2}\right)}+\frac{1}{5\left(\sqrt{2}+\sqrt{3}\right)}+...+\frac{1}{4023\cdot\left(\sqrt{2011}+\sqrt{2012}\right)}< \frac{2011}{2013}\)
1/CMR: \(\forall n\)lẻ thì \(\left(\left(\frac{1+\sqrt{5}}{2}\right)^n+\left(\frac{1-\sqrt{5}}{2}\right)^n\right)^2\) là số chính phương
2/Cho a,b,c>0 và \(a^2+b^2+c^2\le3.CMR:\)
\(\frac{a}{a^2+2b+1}+\frac{b}{b^2+2c+1}+\frac{c}{c^2+2a+1}\le\frac{1}{2}\)
CMR với mọi số tự nhiên n lớn hơn hoặc bằng 1 thì:
\(\left(1+\frac{1}{1\times3}\right)\left(1+\frac{1}{2\times4}\right)\left(1+\frac{1}{3\times5}\right).......\left(1+\frac{1}{n\times\left(n+2\right)}\right)< 2\)