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![](https://rs.olm.vn/images/avt/0.png?1311)
Áp dụng công thức \(\frac{1}{a+1}+\frac{1}{a}< 1\)
\(\frac{1}{5}+\frac{1}{6}< 1;\frac{1}{6}+\frac{1}{7}< 1;...;\frac{1}{16}+\frac{1}{17}< 1\)
ta có: \(\frac{1}{5}+\frac{1}{6}+\frac{1}{7}+...+\frac{1}{17}< 1-\frac{1}{17}\)
\(\Rightarrow\frac{1}{5}+\frac{1}{6}+\frac{1}{7}+...+\frac{1}{17}< 1\)
mà 1<2
\(\Rightarrow\frac{1}{5}+\frac{1}{6}+\frac{1}{7}+...+\frac{1}{17}< 2\)
tham khảo nha bn!
![](https://rs.olm.vn/images/avt/0.png?1311)
\(M=\frac{17}{5}\cdot\frac{-31}{125}\cdot\frac{1}{2}\cdot\frac{10}{17}\cdot\frac{-1}{2^3}\)
\(M=\frac{17}{5}\cdot\frac{-31}{125}\cdot\frac{1}{2}\cdot\frac{10}{7}\cdot\frac{-1}{8}\)
\(M=\left(\frac{17}{5}\cdot\frac{10}{17}\cdot\frac{1}{2}\right)\cdot\frac{-31}{125}\cdot\frac{-1}{8}\)
\(M=1\cdot\frac{31}{1000}=\frac{31}{1000}\)
\(P=\frac{6}{7}\cdot\frac{8}{13}+\frac{6}{9}\cdot\frac{9}{7}-\frac{3}{13}\cdot\frac{6}{7}=\frac{6}{7}\cdot\frac{8}{13}+\frac{6}{7}\cdot1-\frac{3}{13}\cdot\frac{6}{7}\)
\(=\frac{6}{7}\left(\frac{8}{13}+1-\frac{3}{13}\right)=\frac{6}{7}\left(\frac{8}{3}+\frac{13}{13}-\frac{3}{13}\right)=\frac{6}{7}\cdot\frac{18}{13}=\frac{108}{91}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(\frac{5}{17}\left(3\frac{1}{7}+8\frac{7}{3}\right)\frac{15}{17}\left(3\frac{7}{6}+8\frac{1}{7}\right)=\frac{5}{17}\left(\frac{22}{7}+\frac{31}{3}\right)\frac{15}{17}\left(\frac{25}{6}+\frac{57}{7}\right)\)
\(=\frac{5}{17}.\frac{283}{21}.\frac{15}{17}.\frac{517}{42}=\frac{10973325}{254898}=\frac{3657775}{84966}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(a)\frac{1}{3}+\frac{-2}{5}+\frac{1}{6}+\frac{-1}{5}\le x< \frac{-3}{4}+\frac{2}{7}+\frac{-1}{4}+\frac{3}{5}+\frac{5}{7}\)
\(\Rightarrow\frac{1}{3}+\frac{1}{6}+\frac{-2}{5}+\frac{-1}{5}\le x< \frac{-3}{4}+\frac{-1}{4}+\frac{2}{7}+\frac{5}{7}+\frac{3}{5}\)
\(\Rightarrow\frac{2}{6}+\frac{1}{6}+\frac{-3}{5}\le x< -1+1+\frac{3}{5}\)
\(\Rightarrow\frac{1}{2}+\frac{-3}{5}\le x< \frac{3}{5}\)
\(\Rightarrow\frac{-1}{10}\le x< \frac{6}{10}\)
\(\Rightarrow-1\le x< 6\)
\(\Rightarrow x\in\left\{-1;0;1;2;3;4;5\right\}\)
Bài b tương tự
![](https://rs.olm.vn/images/avt/0.png?1311)
Ta có:
\(\frac{1}{5}+\frac{1}{6}+\frac{1}{7}+\frac{1}{8}+\frac{1}{9}< \frac{1}{5}.5=1\) (1)
\(\frac{1}{10}+\frac{1}{11}+...+\frac{1}{16}+\frac{1}{17}< \frac{1}{8}.8=1\) (2)
Cộng theo từng vế (1)và (2)
Ta được:
\(\frac{1}{5}+\frac{1}{6}+...+\frac{1}{17}< 2\)
đầu tiên bạn tách tổng ra là hai 1 la từ 1/5 đến 1/9 còn lại tính rồi ss vs sao lớn nhất trong tổng đẫ tách ra đó
1/143 < 2