\(N=\sqrt{1+2011^2+\frac{2011^2}{2012^2}+\frac{2011}{2012}}\)CMR : N la so tu nhien
Những câu hỏi liên quan
\(S=\sqrt{1+2010^2+\frac{2010^2}{2011^2}}+\frac{2010}{2011}+\sqrt{1+2011^2+\frac{2011^2}{2012^2}}+\frac{2011}{2012}+\sqrt{1+2012^2+\frac{2012^2}{2013^2}}+\frac{2012}{2013}\)
So sánh P và Q biết : P = 2010/2011 + 2011/2012 + 2012/2013 và Q = 2010+2011+2012/ 2011 +2012+2013
Chứng tỏ N < 1 với N = \(\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{2009^2}+\frac{1}{2010^2}\)
Ta có: \(\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{2010^2}
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\(A=\frac{2011}{\sqrt{2012}}+\frac{2012}{\sqrt{2011}};B=\sqrt{2011}+\sqrt{2012}.\)
So sánh A và B
CMR biểu thức \(N=\sqrt{1+2011^2+\dfrac{2011^2}{2012^2}}+\dfrac{2011}{2012}\) có giá trị là 1 số tự nhiên
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.
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So sánh 2 số sau: M=\(\frac{2013^{2012}+2012}{2013^{2011}+1}\)và \(N=\frac{2013^{2011}+2012}{2013^{2010}+1}\)
Ta có :
\(\frac{1}{2013}M=\frac{2013^{2012}+2012}{2013^{2012}+2013}=\frac{2013^{2012}+2013}{2013^{2012}+2013}-\frac{1}{2013^{2012}+2013}=1-\frac{1}{2013^{2012}+2013}\)
Lại có :
\(\frac{1}{2013}N=\frac{2013^{2011}+2012}{2013^{2011}+2013}=\frac{2013^{2011}+2013}{2013^{2011}+2013}-\frac{1}{2013^{2011}+2013}=1-\frac{1}{2013^{2011}+2013}\)
Vì \(\frac{1}{2013^{2012}+2013}< \frac{1}{2013^{2011}+2013}\) nên \(M=1-\frac{1}{2013^{2012}}>N=1-\frac{1}{2013^{2011}+2013}\)
Vậy \(M>N\)
Chúc bạn học tốt ~
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So sánh
M= \(\frac{2010}{2011}+\frac{2011}{2012}+\frac{2012}{2013}\)
N= \(\frac{2010+2011+2012}{2011+2012+2013}\)
N =\(\frac{2010+2011+2012}{2011+2012+2013}\)
\(\Rightarrow N=\frac{2010}{2011+2012+2013}+\frac{2011}{2011+2012+2013}+\frac{2012}{2011+2012+2013}\)
Do: \(\frac{2010}{2011}>\frac{2010}{2011+2012+2013};\frac{2011}{2012}>\frac{2011}{2011+2012+2013};\frac{2012}{2013}>\frac{2012}{2011+2012+2013}\)
\(\Rightarrow\frac{2010}{2011}+\frac{2011}{2012}+\frac{2012}{2013}>\frac{2010}{2011+2012+2013}+\frac{2011}{2011+2012+2013}+\frac{2012}{2011+2012+2013}\)
\(\Rightarrow\frac{2010}{2011}+\frac{2011}{2012}+\frac{2012}{2013}>\frac{2010+2011+2012}{2011+2012+2013}\Leftrightarrow N>M\)
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a=\(\sqrt{1+^{ }2011}^2+^{^{^{\frac{2011^2}{2012^2}}}+\frac{2011}{2012}}\)
tính A biết: \(a+\frac{1}{a+\frac{1}{a+\frac{1}{a+\frac{1}{a}}}}\)
Tính C= \(\frac{1}{1\sqrt{2}+2\sqrt{1}}+\frac{1}{2\sqrt{3}+3\sqrt{2}}+...............+\frac{1}{2011\sqrt{2012}+2012\sqrt{2011}}\)
Ta có: \(\frac{1}{n\sqrt{n+1}+\left(n+1\right)\sqrt{n}}=\frac{1}{\sqrt{n}.\sqrt{n+1}\left(\sqrt{n}+\sqrt{n+1}\right)}\)
\(=\frac{\sqrt{n+1}-\sqrt{n}}{\sqrt{n}.\sqrt{n+1}\left(\sqrt{n}+\sqrt{n+1}\right)\left(\sqrt{n+1}-\sqrt{n}\right)}=\frac{\sqrt{n+1}-\sqrt{n}}{\sqrt{n}.\sqrt{n+1}}\)
\(=\frac{\sqrt{n+1}}{\sqrt{n}.\sqrt{n+1}}-\frac{\sqrt{n}}{\sqrt{n}.\sqrt{n+1}}=\frac{1}{\sqrt{n}}-\frac{1}{\sqrt{n+1}}\)
Thay n = 1, 2, 3, ..., 2011 vào C ta có:
\(C=1-\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{2}}-\frac{1}{\sqrt{3}}+...+\frac{1}{\sqrt{2011}}-\frac{1}{\sqrt{2012}}=1-\frac{1}{\sqrt{2012}}\)
Vậy \(C=1-\frac{1}{\sqrt{2012}}.\)
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uk xie xie (cảm ơn ) bạn , nhưng mik giải ra lâu r
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\(\frac{2012^{2010}+1}{2012^{2011}+1}và\frac{2012^{2011}+1}{2012^{2012}+1}\)so sánh 2 số