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27 tháng 9 2020
ĐKXĐ : \(\left\{{}\begin{matrix}x\ge2011\\y\ge2012\\z\ge2013\end{matrix}\right.\)
Đặt \(\left\{{}\begin{matrix}a=\sqrt{x-2011}\ge0\\b=\sqrt{y-2012}\ge0\\c=\sqrt{z-2013}\ge0\end{matrix}\right.\) ta có :
\(\frac{a-1}{a^2}+\frac{b-1}{b^2}+\frac{c-1}{c^2}=\frac{3}{4}\)
\(\Leftrightarrow\frac{1}{a^2}-\frac{1}{a}+\frac{1}{4}+\frac{1}{b^2}-\frac{1}{b}+\frac{1}{4}+\frac{1}{c^2}-\frac{1}{c}+\frac{1}{4}=0\)
\(\Leftrightarrow\left(\frac{1}{a}-\frac{1}{2}\right)^2+\left(\frac{1}{b}-\frac{1}{2}\right)^2+\left(\frac{1}{c}-\frac{1}{2}\right)^2=0\)
\(\Leftrightarrow a=b=c=2\Leftrightarrow\left\{{}\begin{matrix}x=2015\\y=2016\\z=2017\end{matrix}\right.\)
NN
2
26 tháng 10 2015
\(P=\frac{1}{1.2.3}+\frac{1}{2.3.4}+...+\frac{1}{2011.2012.2013}\)
\(2P=\frac{2}{1.2.3}+\frac{2}{2.3.4}+...+\frac{2}{2011.2012.2013}\)
\(=\left(\frac{1}{1.2}-\frac{1}{2.3}\right)+\left(\frac{1}{2.3}-\frac{1}{3.4}\right)+...+\left(\frac{1}{2011.2012}-\frac{1}{2012.2013}\right)\)
\(=\frac{1}{1.2}-\frac{1}{2012.2013}=\frac{2013.2012-2}{2025078}=\frac{4050154}{2025078}\)
\(\Rightarrow P=\frac{4050154}{2025078}:2\)
26 tháng 10 2015
oh mình tính sai :
\(2P=\frac{1}{1.2}-\frac{1}{2013.2012}=\frac{2013.2012-2025078}{2013.2012}=\frac{4050156-2025078}{4050156}=\frac{2025078}{4050156}\)
\(\Rightarrow P=\frac{2025078}{4050156}:2=\frac{1}{4}\)
NN
2
15 tháng 8 2017
\(\frac{x+1}{2014}+\frac{x+2}{2013}=\frac{x+3}{2012}+\frac{x+4}{2011}\)
\(\left(\frac{x+1}{2014}+1\right)+\left(\frac{x+2}{2013}+1\right)=\left(\frac{x+3}{2012}+1\right)+\left(\frac{x+4}{2011}+1\right)\)
\(\frac{x+2015}{2014}+\frac{x+2015}{2013}=\frac{x+2015}{2012}+\frac{x+2015}{2011}\)
\(\frac{x+2015}{2014}+\frac{x+2015}{2013}-\frac{x+2015}{2012}-\frac{x+2015}{2011}=0\)
\(\left(x+2015\right).\left(\frac{1}{2014}+\frac{1}{2013}-\frac{1}{2012}-\frac{1}{2011}\right)=0\)
\(\Rightarrow\hept{\begin{cases}x+2015=0\\\frac{1}{2014}+\frac{1}{2013}-\frac{1}{2012}-\frac{1}{2011}=0\end{cases}}\)
Vì \(\frac{1}{2014}+\frac{1}{2013}-\frac{1}{2012}-\frac{1}{2011}\ne0\Rightarrow x+2015=0\Rightarrow x=-2015\)
Vậy x = 2015 nha bn
TN
15 tháng 8 2017
\(\frac{x+1}{2014}+\frac{x+2}{2013}=\frac{x+3}{2012}+\frac{x+4}{2011}\)
\(\Rightarrow\left(\frac{x+1}{2014}+1\right)+\left(\frac{x+2}{2013}+1\right)=\left(\frac{x+3}{2012}+1\right)+\left(\frac{x+4}{2011}+1\right)\)
\(\Rightarrow\frac{x+2015}{2014}+\frac{x+2015}{2013}=\frac{x+2015}{2012}+\frac{x+2015}{x+2011}\)
\(\Rightarrow\frac{x+2015}{2014}+\frac{x+2015}{2013}-\frac{x+2015}{2012}-\frac{x+2015}{2011}=0\)
\(\Rightarrow\left(x+2015\right).\left(\frac{1}{2014}+\frac{1}{2013}-\frac{1}{2012}-\frac{1}{2011}\right)\)
Vì \(\frac{1}{2014}+\frac{1}{2013}-\frac{1}{2012}-\frac{1}{2011}\ne0\Rightarrow\left(x-2015\right)=0\)
\(\Rightarrow x=0+2015\) =2015
Đúng thì k ủng hộ mik nha mn!
21 tháng 11 2016
Điều kiện \(\hept{\begin{cases}x-2011>0\\y-2012>0\\z-2013>0\end{cases}\Leftrightarrow\hept{\begin{cases}x>2011\\y>2012\\z>2013\end{cases}}}\)
\(\frac{\sqrt{x-2011}-1}{x-2011}+\frac{\sqrt{y-2012}-1}{y-2012}+\frac{\sqrt{z-2013}-1}{z-2013}=\frac{3}{4}\)
\(\Leftrightarrow\frac{1}{\sqrt{x-2011}}-\frac{1}{x-2011}+\frac{1}{\sqrt{y-2012}}-\frac{1}{y-2012}+\frac{1}{\sqrt{z-2013}}-\frac{1}{z-2013}=\frac{3}{4}\)
\(\Leftrightarrow\left(\frac{1}{x-2011}-\frac{1}{\sqrt{x-2011}}+\frac{1}{4}\right)+\left(\frac{1}{y-2012}-\frac{1}{\sqrt{y-2012}}+\frac{1}{4}\right)+\left(\frac{1}{z-2013}-\frac{1}{\sqrt{z-2013}}+\frac{1}{4}\right)=0\)
\(\Leftrightarrow\left(\frac{1}{\sqrt{x-2011}}-\frac{1}{4}\right)^2+\left(\frac{1}{\sqrt{y-2012}}-\frac{1}{4}\right)^2+\left(\frac{1}{\sqrt{z-2013}}-\frac{1}{4}\right)^2=0\)
\(\Leftrightarrow\hept{\begin{cases}\frac{1}{\sqrt{x-2011}}=\frac{1}{4}\\\frac{1}{\sqrt{y-2012}}=\frac{1}{4}\\\frac{1}{\sqrt{z-2013}}=\frac{1}{4}\end{cases}}\)
\(\Leftrightarrow\hept{\begin{cases}x-2011=16\\y-2012=16\\z-2013=16\end{cases}\Leftrightarrow\hept{\begin{cases}x=2027\\y=2028\\z=2029\end{cases}}}\)
Y
13 tháng 6 2019
2.+ \(\left(2n+1\right)^2=4n^2+4n+1>4n^2+4n\)
\(\Rightarrow2n+1>\sqrt{4n\left(n+1\right)}=2\sqrt{n\left(n+1\right)}\)
+ \(\frac{1}{\left(2n+1\right)\left(\sqrt{n}+\sqrt{n+1}\right)}=\frac{\left(\sqrt{n+1}-\sqrt{n}\right)\left(\sqrt{n+1}+\sqrt{n}\right)}{\left(2n+1\right)\left(\sqrt{n+1}+\sqrt{n}\right)}\)
\(=\frac{\sqrt{n+1}-\sqrt{n}}{2n+1}< \frac{\sqrt{n+1}-\sqrt{n}}{2\sqrt{n\left(n+1\right)}}=\frac{1}{2}\left(\frac{1}{\sqrt{n}}-\frac{1}{\sqrt{n+1}}\right)\)
Do đó : \(A< \frac{1}{2}\left(1-\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{2}}-\frac{1}{\sqrt{3}}+...+\frac{1}{\sqrt{48}}-\frac{1}{\sqrt{49}}\right)\)
\(\Rightarrow A< \frac{1}{2}\)
Y
13 tháng 6 2019
1. + \(\frac{1}{\left(n+1\right)\sqrt{n}}=\frac{\left(n+1\right)-n}{\left(n+1\right)\sqrt{n}}=\frac{\left(\sqrt{n+1}-\sqrt{n}\right)\left(\sqrt{n+1}+\sqrt{n}\right)}{\left(n+1\right)\sqrt{n}}\)
\(< \frac{\left(\sqrt{n+1}-\sqrt{n}\right)\cdot2\sqrt{n+1}}{\sqrt{n}\left(n+1\right)}=2\cdot\frac{n+1-\sqrt{n\left(n+1\right)}}{\left(n+1\right)\sqrt{n}}=2\left(\frac{1}{\sqrt{n}}-\frac{1}{\sqrt{n+1}}\right)\)
Do đó : \(A< 2\left(1-\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{2}}-\frac{1}{\sqrt{3}}+...+\frac{1}{\sqrt{2012}}-\frac{1}{\sqrt{2013}}\right)\)
\(\Rightarrow A< 2\)
Bài 2 tạm thời chưa nghĩ ra :))
Ta có :
42011+42012+42013
= 42011.(1+4+42)
= 42011.21
= 42010.4.21
= 42010.84 ⋮ 84