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\(\Leftrightarrow n+1\in\left\{1;3\right\}\)
hay \(n\in\left\{0;2\right\}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Ta có:
\(\dfrac{1}{5}>\dfrac{1}{10}\\ \dfrac{1}{6}>\dfrac{1}{10}\\ ...\\ \dfrac{1}{9}>\dfrac{1}{10}\\ \Rightarrow\dfrac{1}{5}+\dfrac{1}{6}+...+\dfrac{1}{9}>\dfrac{5}{10}=\dfrac{1}{2}.\)
Tương tự:
\(\dfrac{1}{10}+\dfrac{1}{11}+...+\dfrac{1}{14}>\dfrac{5}{15}=\dfrac{1}{3}.\\ \dfrac{1}{15}+\dfrac{1}{16}+\dfrac{1}{17}>\dfrac{3}{18}=\dfrac{1}{6}.\)
Cộng vế theo vế ta được \(B>\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{6}=1\left(đpcm\right)\)
![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(a,x^3-1=-28\\ \Leftrightarrow x^3=-27\\ \Leftrightarrow x^3=\left(-3\right)^3\\ \Leftrightarrow x=-3\\ b,\left(y-1\right)^2-32=-23\\ \Leftrightarrow\left(y-1\right)^2=9\\ \Leftrightarrow\left[{}\begin{matrix}y-1=3\\y-1=-3\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}y=4\\y=-2\end{matrix}\right.\\ c,15-16:\left|x\right|=-1\\ \Leftrightarrow16:\left|x\right|=16\\ \Leftrightarrow\left|x\right|=1\\ \Leftrightarrow x=\pm1\)
![](https://rs.olm.vn/images/avt/0.png?1311)
A<1/1*2+1/2*3+...+1/2021*2022
=>A<1-1/2+1/2-1/3+...+1/2021-1/2022<1
![](https://rs.olm.vn/images/avt/0.png?1311)
Đặt \(A=\dfrac{10^{1987}+1}{10^{1988}+1};B=\dfrac{10^{1989}+1}{10^{1990}+1}\)
Ta có: \(A=\dfrac{10^{1987}+1}{10^{1988}+1}\)
\(\Leftrightarrow10A=\dfrac{10^{1988}+10}{10^{1988}+1}=1+\dfrac{9}{10^{1988}+1}\)
Ta có: \(B=\dfrac{10^{1989}+1}{10^{1990}+1}\)
\(\Leftrightarrow10B=\dfrac{10^{1990}+10}{10^{1990}+1}=1+\dfrac{9}{10^{1990}+1}\)
Ta có: \(10^{1988}+1< 10^{1990}+1\)
\(\Leftrightarrow\dfrac{9}{10^{1988}+1}>\dfrac{9}{10^{1990}+1}\)
\(\Leftrightarrow1+\dfrac{9}{10^{1988}+1}>1+\dfrac{9}{10^{1990}+1}\)
hay A>B