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![](https://rs.olm.vn/images/avt/0.png?1311)
Xem lại cái đề là \(\frac{2}{7.11}\)hay \(\frac{2}{11}\)nha
![](https://rs.olm.vn/images/avt/0.png?1311)
\(\frac{3}{4}+\frac{1}{4}:X=\frac{1}{2}\)
=> \(\frac{1}{4}:X=\frac{1}{2}-\frac{3}{4}\)
=> \(\frac{1}{4}:X=\frac{4}{8}-\frac{6}{8}\)
=> \(\frac{1}{4}:X=\frac{-1}{4}\)
=>\(X=\frac{1}{4}:\frac{-1}{4}\)
=> X = -1
![](https://rs.olm.vn/images/avt/0.png?1311)
Sửa đề: \(\frac{5}{2.1}+\frac{4}{1.11}+\frac{3}{11.2}+\frac{1}{2.15}+\frac{13}{15.4}\)
Đặt B là tên của biểu thức
\(B=\frac{5}{2.1}+\frac{4}{1.11}+\frac{3}{11.2}+\frac{1}{2.15}+\frac{13}{15.4}\)
\(\frac{1}{7}B=\frac{5}{2.7}+\frac{4}{7.11}+\frac{3}{11.14}+\frac{1}{14.15}+\frac{13}{15.28}\)
\(\frac{1}{7}B=\frac{1}{2}-\frac{1}{7}+\frac{1}{7}-\frac{1}{11}+\frac{1}{11}-\frac{1}{14}+\frac{1}{14}-\frac{1}{15}+\frac{1}{15}-\frac{1}{28}\)
\(\frac{1}{7}B=\frac{1}{2}-\frac{1}{28}=\frac{13}{28}\)
\(B=\frac{13}{28}:\frac{1}{7}=\frac{13}{4}=3\frac{1}{4}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
a)\(19\frac{5}{8}:\frac{7}{12}-15\frac{1}{4}:\frac{7}{12}\)
=(\(19\frac{5}{8}-15\frac{1}{4}\)):\(\frac{7}{12}\)
=(\(19\frac{10}{16}-15\frac{4}{16}\)):\(\frac{7}{12}\)
=\(4\frac{6}{16}:\frac{7}{12}\)
=\(\frac{35}{8}:\frac{7}{12}\)
=\(\frac{35}{8}\cdot\frac{12}{7}\)
=\(\frac{15}{2}\)
b)2/5*1/3-2/15:1/5+3/5*1/3
=2/15-2/3+1/5
=-8/15+1/5
=-1/3
aidi qua dong tinh nho h chom minh nhe
![](https://rs.olm.vn/images/avt/0.png?1311)
Xét TH1 : ( S < 8/9 )
\(\frac{1}{2\cdot2}< \frac{1}{1\cdot2};\frac{1}{3\cdot3}< \frac{1}{2\cdot3};...;\frac{1}{9\cdot9}< \frac{1}{8\cdot9}\)
\(\Rightarrow\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{9^2}< \frac{1}{1\cdot2}+\frac{1}{2\cdot3}+...+\frac{1}{8\cdot9}\)
hay \(S< \frac{1}{1\cdot2}+\frac{1}{2\cdot3}+...+\frac{1}{8\cdot9}\)
\(S< 1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{8}-\frac{1}{9}\)
\(S< 1-\frac{1}{9}=\frac{8}{9}\left(1\right)\)
TH2 : ( S > 2/5 )
\(\frac{1}{2\cdot2}>\frac{1}{2\cdot3};\frac{1}{3\cdot3}>\frac{1}{3\cdot4};...;\frac{1}{9\cdot9}>\frac{1}{9\cdot10}\)
\(\Rightarrow\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{9^2}>\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+...+\frac{1}{9\cdot10}\)
hay \(S>\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+...+\frac{1}{9\cdot10}\)
\(S>\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{9}-\frac{1}{10}\)
\(S>\frac{1}{2}-\frac{1}{10}=\frac{2}{5}\left(2\right)\)
Từ (1) và (2) => đpcm
Ko tk thì ko phải là ng` nx rồi :)
![](https://rs.olm.vn/images/avt/0.png?1311)
a)\(\frac{2}{3}+\frac{3}{4}+\frac{5}{6}\)
\(=\frac{8+9+10}{12}\)
\(=\frac{27}{12}=\frac{9}{4}\)
b)\(\frac{15}{8}-\frac{7}{12}+\frac{5}{6}\)
\(=\frac{45-14+20}{24}\)
\(=\frac{51}{24}=\frac{17}{8}\)
2)
a)\(\frac{2}{5}+\frac{7}{13}+\frac{3}{5}+\frac{1}{7}\)
\(=\frac{2}{5}+\frac{3}{5}+\frac{7}{13}+\frac{1}{7}\)
\(=1+\frac{7}{13}+\frac{1}{7}\)
\(=\frac{20}{13}+\frac{1}{7}\)
\(=\frac{153}{91}\)
Tí tớ trả lời tiếp