1+1+1-1-1+2+55
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![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
a) Ta có: \(\dfrac{x}{14}-\dfrac{1}{7}=\dfrac{-3}{4}\)
\(\Leftrightarrow\dfrac{x}{14}=\dfrac{-3}{4}+\dfrac{1}{7}=\dfrac{-21}{28}+\dfrac{4}{28}=\dfrac{-17}{28}\)
hay \(x=\dfrac{-17\cdot14}{28}=\dfrac{-17}{2}\)
Vậy: \(x=-\dfrac{17}{2}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(\dfrac{1}{55^2}.5^4.\left(-11\right)^2.55^5.\left(\dfrac{1}{5^2}\right)^2:\left(5^3.11^6\right)\)
\(=\dfrac{5^4.11^2.5^5.11^5}{5^2.11^2}.\dfrac{1}{5^4.5^3.11^6}\)
\(=\dfrac{5^9.11^7}{5^9.11^8}=\dfrac{1}{11}\)
\(=\dfrac{1}{5^2\cdot11^2}\cdot5^4\cdot11^2\cdot55^5\cdot\left(\dfrac{1}{5}\right)^2\cdot\dfrac{1}{5^3\cdot11^6}\)
\(=5^2\cdot5^5\cdot11^5\cdot\dfrac{1}{5^2\cdot5^3\cdot11^6}\)
\(=\dfrac{11^5}{11^6}\cdot\dfrac{5^7}{5^5}=\dfrac{25}{11}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
a)
\(\dfrac{1}{2\cdot3}x+\dfrac{1}{3\cdot4}x+...+\dfrac{1}{49\cdot50}x=1\\ x\left(\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{49}-\dfrac{1}{50}\right)=1\\ x\left(\dfrac{1}{2}-\dfrac{1}{50}\right)=1\\ x\cdot\dfrac{12}{25}=1\\ x=1:\dfrac{12}{25}=1\cdot\dfrac{25}{12}=\dfrac{25}{12}\)
=59 nha
=58 nha
k cho mk nha
kb voi mk nha