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\(A=\sqrt{x^2-2x+1}+\sqrt{x^2+4x+4}\)
\(=\sqrt{\left(x-1\right)^2}+\sqrt{\left(x+2\right)^2}\)
\(=|1-x|+|x+2|\ge|1-x+x+2|=3\)
\(x\sqrt{x+\frac{1}{2}+\sqrt{x+\frac{1}{4}}}=2\)
\(\Leftrightarrow x\sqrt{\left(\sqrt{x+\frac{1}{4}}+\frac{1}{2}\right)^2}=2\)
\(\Leftrightarrow x\sqrt{x+\frac{1}{4}}+\frac{1}{2}=2\)
\(\Leftrightarrow x\sqrt{x+\frac{1}{4}}=\frac{3}{2}\)
Làm nốt
Đặt: \(\hept{\begin{cases}\sqrt{x-2009}=a\\\sqrt{y-2010}=b\\\sqrt{z-2011}=c\end{cases}}\)
Ta có: \(\frac{1}{a}-\frac{1}{a^2}+\frac{1}{b}-\frac{1}{b^2}+\frac{1}{c}-\frac{1}{c^2}-\frac{3}{4}=0\)
\(\Leftrightarrow\frac{1}{a^2}-\frac{1}{a}+\frac{1}{b^2}-\frac{1}{b}+\frac{1}{c^2}-\frac{1}{c}+\frac{3}{4}=0\)
\(\Leftrightarrow\left(\frac{1}{a^2}-\frac{1}{a}+\frac{1}{4}\right)+\left(\frac{1}{b^2}-\frac{1}{b}+\frac{1}{4}\right)+\left(\frac{1}{c^2}-\frac{1}{c}+\frac{1}{4}\right)=0\)
\(\Leftrightarrow\left(\frac{1}{a}-\frac{1}{2}\right)^2+\left(\frac{1}{b}-\frac{1}{2}\right)^2+\left(\frac{1}{c}-\frac{1}{2}\right)^2=0\)
\(\Leftrightarrow a=b=c=\frac{1}{2}\)
Thay vào tìm x;y;z
Đặt: \(\hept{\begin{cases}\sqrt{x-2009}=a\\\sqrt{y-2010}=b\\\sqrt{z-2011}=c\end{cases}}\)
Ta có: \frac{1}{a}-\frac{1}{a^2}+\frac{1}{b}-\frac{1}{b^2}+\frac{1}{c}-\frac{1}{c^2}-\frac{3}{4}=0a1−a21+b1−b21+c1−c21−43=0
\Leftrightarrow\frac{1}{a^2}-\frac{1}{a}+\frac{1}{b^2}-\frac{1}{b}+\frac{1}{c^2}-\frac{1}{c}+\frac{3}{4}=0⇔a21−a1+b21−b1+c21−c1+43=0
\Leftrightarrow\left(\frac{1}{a^2}-\frac{1}{a}+\frac{1}{4}\right)+\left(\frac{1}{b^2}-\frac{1}{b}+\frac{1}{4}\right)+\left(\frac{1}{c^2}-\frac{1}{c}+\frac{1}{4}\right)=0⇔(a21−a1+41)+(b21−b1+41)+(c21−c1+41)=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⇔(a1−21)2+(b1−21)2+(c1−21)2=0
\Leftrightarrow a=b=c=\frac{1}{2}⇔a=b=c=21
Thay vào tìm x;y;z
\(\left(2n+1\right)^2=4n^2+4n+1\)
\(>4n^2+4n=4n\left(n+1\right)\)
\(\Rightarrow2n+1>\sqrt{4n\left(n+1\right)}=2\sqrt{n\left(n+1\right)}\)
\(\Rightarrow\frac{\sqrt{n+1}-\sqrt{n}}{2n+1}< \frac{1}{2}\cdot\frac{\sqrt{n+1}-\sqrt{n}}{\sqrt{n\left(n+1\right)}}\) \(=\frac{1}{2}\cdot\left(\frac{1}{\sqrt{n}}-\frac{1}{\sqrt{n+1}}\right)\)
Do đó : \(\frac{\sqrt{2}-\sqrt{1}}{3}+\frac{\sqrt{3}-\sqrt{2}}{5}+...+\frac{\sqrt{2011}-\sqrt{2010}}{4021}\)
\(< \frac{1}{2}\cdot\left(\frac{1}{\sqrt{1}}-\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{2}}-\frac{1}{\sqrt{3}}+...+\frac{1}{\sqrt{2010}}-\frac{1}{\sqrt{2011}}\right)\)
\(< \frac{1}{2}\)
Xét:
\(1+\frac{1}{k^2}+\frac{1}{\left(k+1\right)^2}=\left(\frac{1}{k}-\frac{1}{k+1}\right)^2+\frac{2}{k\left(k+1\right)}+1=\frac{1}{k^2\left(k+1\right)^2}+\frac{2}{k\left(k+1\right)}+1=\left(\frac{1}{k\left(k+1\right)}+1\right)^2\)
\(\Rightarrow\sqrt{1+\frac{1}{k^2}+\frac{1}{\left(k+1\right)^2}}=\frac{1}{k\left(k+1\right)}+1\)\(=1+\frac{1}{k}-\frac{1}{k+1}\)
Cho \(k\)chạy từ 1 đến 2010 ta có
Tổng cần tính
\(=\)\(1+1-\frac{1}{2}+1+\frac{1}{2}-\frac{1}{3}+1+\frac{1}{3}-\frac{1}{4}+...+1+\frac{1}{2010}-\frac{1}{2011}\)
\(=2011-\frac{1}{2011}=\frac{2010.2012}{2011}\)
Thưa bn mk đã làm ra nhưng không biết có đúng không. Xem nhá:
Ta có:
\(\frac{\sqrt{x-2009}-1}{x-2009}+\frac{\sqrt{y-2001}-1}{y-2001}+\frac{\sqrt{z-2011}-1}{z-2011}=\frac{3}{4}\Leftrightarrow"\frac{1}{\sqrt{x-2009}}-\frac{1}{2}"^2+\)
\("\frac{1}{\sqrt{y-2010}}-\frac{1}{2}"^2-"\frac{1}{\sqrt{z-2011}}-\frac{1}{2}"^2=0\)
\(\Rightarrow x=2013;y=2014;z=2015\)
P/s: Bn thay Ngoặc Kép thành Ngoặc Đơn nhé
(1 +2010) > 2\(\sqrt{1.2010}\)=> \(\frac{1}{\sqrt{1.2010}}\)> 2/2011 tương tự các phần tử còn lại
vậy C > 2/2011+2/2011+.....2/2011 = 2.2010/2011