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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
\(\left[\frac{\left(1-\sqrt{x}\right)\left(x+\sqrt{x}+1\right)}{1-\sqrt{x}}\right]\left[\frac{1-\sqrt{x}}{\left(1-\sqrt{x}\right)\left(1+\sqrt{x}\right)}\right]^2=\left(x+\sqrt{x}+1\right)\frac{1}{\left(1+\sqrt{x}\right)^2}=\frac{x+\sqrt{x}+1}{x+2\sqrt{x}+1}\)
Đề bài sai
\(\sqrt{2012}-\sqrt{2011}=\frac{1}{\sqrt{2012}+\sqrt{2011}}\)
\(\sqrt{2011}-\sqrt{2010}=\frac{1}{\sqrt{2011}+\sqrt{2010}}\)
Do \(\sqrt{2012}>\sqrt{2010}\) \(\Rightarrow\sqrt{2012}+\sqrt{2011}>\sqrt{2011}+\sqrt{2010}>0\)
\(\Rightarrow\frac{1}{\sqrt{2012}+\sqrt{2011}}< \frac{1}{\sqrt{2011}+\sqrt{2010}}\Rightarrow\sqrt{2012}-\sqrt{2011}< \sqrt{2011}-\sqrt{2010}\)
\(A=\frac{x+2\sqrt{xy}+y-4\sqrt{xy}}{\sqrt{x}-\sqrt{y}}+\frac{\sqrt{xy}\left(\sqrt{x}-\sqrt{y}\right)}{\sqrt{xy}}=\frac{\left(\sqrt{x}-\sqrt{y}\right)^2}{\sqrt{x}-\sqrt{y}}+\frac{\sqrt{xy}\left(\sqrt{x}-\sqrt{y}\right)}{\sqrt{xy}}\)
\(=\sqrt{x}-\sqrt{y}+\sqrt{x}-\sqrt{y}=2\sqrt{x}-2\sqrt{y}\)
\(M^2=\left(\sqrt{x-1}+\sqrt{9-x}\right)^2\le2\left(x-1+9-x\right)=16\)
\(\Rightarrow M\le4\Rightarrow M_{max}=4\) khi \(x-1=9-x\Leftrightarrow x=5\)
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 : \(A< \frac{1}{2}\left(1-\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{2}}-\frac{1}{\sqrt{3}}+...+\frac{1}{\sqrt{48}}-\frac{1}{\sqrt{49}}\right)\)
\(\Rightarrow A< \frac{1}{2}\)
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 :))
\(\left(2n+1\right)^2=4n^2+4n+1\)
\(>4n^2+4n=4n\left(n+1\right)\)
\(\Rightarrow2n+1>\sqrt{4n\left(n+1\right)}=2\sqrt{n\left(n+1\right)}\)
\(\Rightarrow\frac{\sqrt{n+1}-\sqrt{n}}{2n+1}< \frac{1}{2}\cdot\frac{\sqrt{n+1}-\sqrt{n}}{\sqrt{n\left(n+1\right)}}\) \(=\frac{1}{2}\cdot\left(\frac{1}{\sqrt{n}}-\frac{1}{\sqrt{n+1}}\right)\)
Do đó : \(\frac{\sqrt{2}-\sqrt{1}}{3}+\frac{\sqrt{3}-\sqrt{2}}{5}+...+\frac{\sqrt{2011}-\sqrt{2010}}{4021}\)
\(< \frac{1}{2}\cdot\left(\frac{1}{\sqrt{1}}-\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{2}}-\frac{1}{\sqrt{3}}+...+\frac{1}{\sqrt{2010}}-\frac{1}{\sqrt{2011}}\right)\)
\(< \frac{1}{2}\)
Đặt B là tên biểu thức
Với mọi n thuộc N*, ta có:
\(\frac{1}{\left(n+1\right)\sqrt{n}}=\frac{\sqrt{n}}{n\left(n+1\right)}=\sqrt{n}\left(\frac{1}{n}-\frac{1}{n+1}\right)=\sqrt{n}\left(\frac{1}{\sqrt{n}}+\frac{1}{\sqrt{n+1}}\right)\left(\frac{1}{\sqrt{n}}-\frac{1}{\sqrt{n+1}}\right)\)
\(=\left(1+\frac{\sqrt{n}}{\sqrt{n+1}}\right)\left(\frac{1}{\sqrt{n}}-\frac{1}{\sqrt{n+1}}\right)< 2\left(\frac{1}{\sqrt{n}}-\frac{1}{\sqrt{n+1}}\right)\) (*)
Áp dụng (*), ta được:
\(B< 2\left(1-\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{2}}-\frac{1}{\sqrt{3}}+\frac{1}{\sqrt{3}}-\frac{1}{\sqrt{4}}+...+\frac{1}{\sqrt{2011}}-\frac{1}{\sqrt{2012}}+\frac{1}{\sqrt{2012}}-\frac{1}{\sqrt{2013}}\right)\)
\(=2\left(1-\frac{1}{\sqrt{2013}}\right)=2-\frac{1}{\sqrt{2013}}< 2\)
(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