2021/1x2+2021/2x3+2021/3x4+2021/4x5+...+2021/9x10
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Ta thấy \(\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}+\dfrac{z^2}{c^2}\ge\dfrac{x^2}{a^2+b^2+c^2}+\dfrac{y^2}{a^2+b^2+c^2}+\dfrac{z^2}{a^2+b^2+c^2}=\dfrac{x^2+y^2+z^2}{a^2+b^2+c^2}\).
Mà đẳng thức xảy ra nên ta phải có x = y = z = 0 (Do \(a^2,b^2,c^2>0\)).
Thay vào đẳng thức cần cm ta có đpcm.
![](https://rs.olm.vn/images/avt/0.png?1311)
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Xét khai triển:
\(2^{2021}=\left(1+1\right)^{2021}=C_{2021}^0+C_{2021}^1+...+C_{2021}^{2020}+C_{2021}^{2021}\) (1)
\(0=\left(1-1\right)^{2021}=C_{2021}^0-C_{2021}^1+C_{2021}^2+...+C_{2021}^{2020}-C_{2021}^{2021}\) (2)
Trừ vế cho vế (1) và (2):
\(2^{2021}=2.C_{2021}^1+2.C_{2021}^3+...+2C_{2021}^{2021}\)
\(\Rightarrow C_{2021}^1+C_{2021}^3+...+C_{2021}^{2019}+C_{2021}^{2021}=\dfrac{2^{2021}}{2}=2^{2020}\)
\(\Rightarrow C_{2021}^1+C_{2021}^3+...+C_{2021}^{2019}+1=2^{2020}\)
\(\Rightarrow C_{2021}^1+C_{2021}^3+...+C_{2021}^{2019}=2^{2020}-1\)
\(\frac{2021}{1\cdot2}+\frac{2021}{2\cdot3}+...+\frac{2021}{9\cdot10}=2021\cdot\left(\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+...+\frac{1}{9\cdot10}\right)\)
\(=2021\cdot\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{9}-\frac{1}{10}\right)=2021\cdot\left(1-\frac{1}{10}\right)\)
\(=2021\cdot\frac{9}{10}=\frac{18189}{10}\)
Ta có : \(\frac{2021}{1.2}+\frac{2021}{2.3}+\frac{2021}{3.4}+...+\frac{2021}{9.10}=2021\left(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{9.10}\right)\)
\(=2021\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{9}-\frac{1}{10}\right)\)
\(=2021\left(1-\frac{1}{10}\right)=2021.\frac{9}{10}=1818,9\)