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![](https://rs.olm.vn/images/avt/0.png?1311)
\(4B< \frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{n\left(n+1\right)}\)
\(4B< \frac{2-1}{1.2}+\frac{3-2}{2.3}+...+\frac{n+1-n}{n\left(n+1\right)}=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{n}-\frac{1}{n+1}\)
\(4B< 1-\frac{1}{n+1}\Rightarrow B< \frac{1}{4}-\frac{1}{4\left(n+1\right)}< \frac{1}{4}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(1,-\frac{3}{29}+\frac{-7}{29}\le\frac{x}{29}\le-\frac{3}{29}-\frac{5}{29}\)
\(\Rightarrow-\frac{10}{29}\le\frac{x}{29}\le-\frac{8}{29}\Rightarrow-10\le x\le-8\)
\(\Rightarrow x=\left\{-8;-9;-10\right\}\)
\(S=\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+\frac{1}{2^4}+...+\frac{1}{2^{99}}+\frac{1}{2^{100}}\)
\(\Rightarrow2S=1+\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{99}}\)
\(S=\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+\frac{1}{2^4}+...+\frac{1}{2^{99}}+\frac{1}{2^{100}}\)
\(\Rightarrow2S-S=S=1-\frac{1}{2^{100}}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Bài làm:
Ta có: \(-\frac{7}{12}\le x\le\frac{1}{4}\)
\(\Leftrightarrow-1< x< 1\)
\(\Rightarrow x=0\)
Vậy x = 0
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![](https://rs.olm.vn/images/avt/0.png?1311)
\(\frac{x}{5}\le\frac{12}{x}\Rightarrow x^2\le60\left(1\right)\)
\(\frac{12}{x}\le\frac{x}{3}\Rightarrow x^2\ge36\left(2\right)\)
Từ (1) và (2)
\(\Rightarrow36\le x^2\le60\) và \(x\in N\)
\(\Rightarrow6\le x\le7,75\)
Vậy \(x=6;7\)
![](https://rs.olm.vn/images/avt/0.png?1311)
a)\(\frac{-1}{2}\le n\le5\left(n\in Z\right)\)
\(\Rightarrow-0,5\le n\le5\)
Mà \(n\in Z\)
\(\Rightarrow n\in\left\{0;1;2;3;4;5\right\}\)
Vậy \(n\in\left\{0;1;2;3;4;5\right\}\)
b)Tương tự phần a
c)\(\frac{-4}{9}< n\le\frac{-1}{2}\left(n\in Z\right)\)
\(\Rightarrow-0,44< n\le-0,5\)
Mà \(n\in Z\)
\(\Rightarrow n\in\varnothing\)
Vậy \(n\in\varnothing\)
Chúc bn học tốt
![](https://rs.olm.vn/images/avt/0.png?1311)
Bạn vào link này nha:https://hoc24.vn/hoi-dap/question/941948.html
Chúc bn học tốt
![](https://rs.olm.vn/images/avt/0.png?1311)
\(\left(\frac{-2}{3}-\frac{1}{2}\right):\frac{-1}{4}\le x\le\left(\frac{-5}{6}+\frac{2}{\frac{1}{4}}:\frac{-3}{2}\right)\cdot\left(\frac{-7}{\frac{1}{2}}\right)\)
\(taco:\left(\frac{-2}{3}-\frac{1}{2}\right):\frac{-1}{4}=\frac{-7}{6}:\frac{-1}{4}=\frac{14}{3}\)
\(\left(\frac{-5}{6}+\frac{2}{\frac{1}{4}}:\frac{-3}{2}\right)\cdot\left(\frac{-7}{\frac{1}{2}}\right)=\left(\frac{-5}{6}+\frac{-16}{3}\right)\cdot\left(-14\right)=\frac{-37}{6}\cdot\left(-14\right)=\frac{259}{3}\)
TU DO \(=>X=\frac{14}{3};\frac{15}{3};,,,;\frac{259}{3}\)
CHUC BAN HOC TOT :))