\(\frac{2}{5}cua25la?\)
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![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(\frac{5}{7}\times\frac{1}{3}-\frac{5}{7}\times\frac{1}{4}-\frac{5}{7}\times\frac{1}{2}\)
\(=\frac{5}{7}\times\left(\frac{1}{3}-\frac{1}{4}-\frac{1}{2}\right)\)
\(=\frac{5}{7}\times\left(\frac{4}{12}-\frac{3}{12}-\frac{6}{12}\right)\)
\(=\frac{5}{7}\times\left(\frac{4-3-6}{12}\right)\)
\(=\frac{5}{7}\times\frac{-5}{12}\)
\(=\frac{5\times\left(-5\right)}{7\times12}\)
\(=\frac{-25}{84}\)
\(\frac{5}{7}.\frac{1}{3}-\frac{5}{7}.\frac{1}{4}-\frac{5}{7}.\frac{1}{2}\)
= \(\frac{5}{7}.\left(\frac{1}{3}-\frac{1}{4}-\frac{1}{2}\right).1\)
\(=\frac{5}{7}.\frac{-5}{12}\)
\(=-\frac{25}{84}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
a)
\(\Rightarrow A=\frac{\frac{1}{11}-\frac{1}{13}-\frac{1}{17}}{5\left(\frac{1}{11}-\frac{1}{13}-\frac{1}{17}\right)}+\frac{2\left(\frac{1}{3}-\frac{1}{9}-\frac{1}{27}+\frac{1}{81}\right)}{7\left(\frac{1}{3}-\frac{1}{9}-\frac{1}{27}+\frac{1}{81}\right)}\)
\(\Rightarrow A=\frac{1}{5}+\frac{2}{7}\)
\(\Rightarrow A=\frac{17}{35}\)
b)
\(\Rightarrow B=5\left(\frac{1}{11}-\frac{1}{16}+\frac{1}{16}-\frac{1}{21}+....+\frac{1}{56}-\frac{1}{61}\right)\)
\(\Rightarrow B=5\left(\frac{1}{11}-\frac{1}{61}\right)\)
\(\Rightarrow B=5.\frac{50}{671}=\frac{250}{671}\)
c)
\(\Rightarrow C=1-\left(\frac{1}{1.3}+\frac{1}{2.3}+\frac{1}{2.5}+....+\frac{1}{49.25}\right)\)
\(\Rightarrow C=1-2\left(\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+....+\frac{1}{49.50}\right)\)
\(\Rightarrow C=1-2\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+....+\frac{1}{49}-\frac{1}{50}\right)\)
\(\Rightarrow C=1-1-\frac{1}{25}\)
\(\Rightarrow C=\frac{1}{25}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
(A) \(\frac{2}{3} + \frac{{ - 4}}{6} = \frac{4}{6} + \frac{{ - 4}}{6} = 0\) => A sai
(B) \(\frac{2}{3}.\frac{{ - 1}}{5} = \frac{{ - 2}}{{15}}\) mà \(\frac{{3 - 2}}{5} = \frac{1}{5}\) => B sai
(C) \(\frac{2}{3} - \frac{3}{5} = \frac{{10}}{{15}} - \frac{9}{{15}} = \frac{1}{{15}}\) => C đúng
(D) \(\frac{3}{5}:\frac{3}{{ - 5}} = \frac{3}{5}.\frac{{ - 5}}{3} = \frac{{ - 15}}{{15}} = - 1\) => D sai
=> Chọn C.
![](https://rs.olm.vn/images/avt/0.png?1311)
Ta có:
\(A=\frac{5^2}{1.6}+\frac{5^2}{6.11}+\frac{5^2}{11.16}+...+\frac{5^2}{26.31}\)
\(A=5\left(\frac{5}{1.6}+\frac{5}{6.11}+...+\frac{5}{26.31}\right)\)
\(A=5\left(\frac{1}{1}-\frac{1}{6}+\frac{1}{6}-\frac{1}{11}+...+\frac{1}{26}-\frac{1}{31}\right)\)
\(A=5\left(\frac{1}{1}-\frac{1}{31}\right)\)
\(A=5.\frac{30}{31}\)
\(A=\frac{150}{31}\)
Vậy \(A=\frac{150}{31}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(\frac{5^2}{1.6}+\frac{5^2}{6.11}+...+\frac{5^2}{26.31}=5\left(\frac{5}{1.6}+\frac{5}{6.11}+...+\frac{5}{26.31}\right)=5\left(\frac{1}{1}-\frac{1}{6}+\frac{1}{6}-...+\frac{1}{26}-\frac{1}{31}\right)\)
\(=5\left(1-\frac{1}{31}\right)=\frac{5.30}{31}=\frac{150}{31}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Câu 1:
Giả sử \(\frac{3}{5}< \frac{3+m}{5+m}\)
=) \(3.\left(5+m\right)< 5.\left(3+m\right)\)
=) \(15+3m< 15+5m\) ( Đúng vì \(15=15\)và \(3m< 5m\)) =) Điều giả sử đúng
=) \(\frac{3}{5}< \frac{3+m}{5+m}\)
* Từ điều trên ta suy ra : Nếu \(\frac{a}{b}< 1\)=) \(\frac{a}{b}< \frac{a+m}{b+m}\)
Và nếu \(\frac{a}{b}>1\)=) \(\frac{a}{b}>\frac{a+m}{b+m}\)
Câu 2 :
= \(5.\left(\frac{5}{1.6}+\frac{5}{6.11}+...+\frac{5}{26.31}\right)\)
= \(5.\left(\frac{1}{1}-\frac{1}{6}+\frac{1}{6}-\frac{1}{11}+...+\frac{1}{26}-\frac{1}{31}\right)\)
= \(5.\left(\frac{1}{1}-\frac{1}{31}\right)\)= \(5.\frac{30}{31}=\frac{150}{31}\)
=> Với mọi số tự nhiên m ( như m\(\ne\)0 ) thì \(\frac{3}{5}< \frac{3+m}{5+m}\)
\(\frac{5^2}{1.6}+\frac{5^2}{6.11}+\frac{5^2}{11.16}+\frac{5^2}{16.21}+\frac{5^2}{21.26}+\frac{5^2}{26.31}\)
\(=5\left(\frac{1}{1.6}+\frac{1}{6.11}+...+\frac{1}{26.31}\right)\)
\(=5\left(1-\frac{1}{6}+\frac{1}{6}-\frac{1}{11}+...+\frac{1}{26}-\frac{1}{31}\right)\)
\(=5\left(1-\frac{1}{31}\right)\)
\(=5.\frac{30}{31}\)
\(=\frac{150}{31}\)
= 10
nha bạn
tick mk