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![](https://rs.olm.vn/images/avt/0.png?1311)
Ta có \(M=\dfrac{1}{1+2+3}+\dfrac{1}{1+2+3+4}+...+\dfrac{1}{1+2+3+...+59}\)
= \(\dfrac{1}{3\cdot4}+\dfrac{1}{4\cdot5}+...+\dfrac{1}{59\cdot60}\)
= \(\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{5}+...+\dfrac{1}{59}-\dfrac{1}{60}\)
= \(\dfrac{1}{3}-\dfrac{1}{60}=\dfrac{19}{60}< \dfrac{40}{60}=\dfrac{2}{3}\)
Vậy M < \(\dfrac{2}{3}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
a) \(\dfrac{3}{7}x-2\dfrac{1}{3}=0,5\)
\(\Leftrightarrow\dfrac{3}{7}x-\dfrac{7}{3}=\dfrac{1}{2}\)
\(\Leftrightarrow\dfrac{3}{7}x=\dfrac{1}{2}+\dfrac{7}{3}\)
\(\Leftrightarrow\dfrac{3}{7}x=\dfrac{17}{6}\)
\(\Leftrightarrow x=\dfrac{17}{6}:\dfrac{3}{7}\)
\(\Leftrightarrow x=\dfrac{119}{18}\)
b) \(\dfrac{4}{7}-\dfrac{2}{3}:x=1\dfrac{1}{4}\)
\(\Leftrightarrow\dfrac{4}{7}-\dfrac{2}{3}:x=\dfrac{5}{4}\)
\(\Leftrightarrow\dfrac{2}{3}:x=\dfrac{4}{7}-\dfrac{5}{4}\)
\(\Leftrightarrow\dfrac{2}{3}:x=\dfrac{-19}{28}\)
\(\Leftrightarrow x=\dfrac{2}{3}:\dfrac{-19}{28}\)
\(\Leftrightarrow x=\dfrac{56}{-57}\)
c) \(\left(\dfrac{2}{3}x+2\dfrac{1}{4}\right):3\dfrac{1}{5}=0,75\)
\(\Leftrightarrow\left(\dfrac{2}{3}x+\dfrac{9}{4}\right):\dfrac{16}{6}=\dfrac{3}{4}\)
\(\Leftrightarrow\dfrac{2}{3}x+\dfrac{9}{4}=\dfrac{3}{4}.\dfrac{16}{6}\)
\(\Leftrightarrow\dfrac{2}{3}x+\dfrac{9}{4}=2\)
\(\Leftrightarrow\dfrac{2}{3}x=2-\dfrac{9}{4}\)
\(\Leftrightarrow\dfrac{2}{3}x=\dfrac{-1}{4}\)
\(\Leftrightarrow x=\dfrac{-1}{4}:\dfrac{2}{3}\)
\(\Leftrightarrow x=\dfrac{-3}{8}\)
d) \(\left|\dfrac{4}{5}-\dfrac{2}{3}x\right|:\dfrac{1}{4}-\dfrac{2}{3}=1\)
\(\Leftrightarrow\left|\dfrac{4}{5}-\dfrac{2}{3}x\right|:\dfrac{1}{4}=1+\dfrac{2}{3}\)
\(\Leftrightarrow\left|\dfrac{4}{5}-\dfrac{2}{3}x\right|:\dfrac{1}{4}=\dfrac{5}{3}\)
\(\Leftrightarrow\left|\dfrac{4}{5}-\dfrac{2}{3}x\right|=\dfrac{5}{3}.\dfrac{1}{4}\)
\(\Leftrightarrow\left|\dfrac{4}{5}-\dfrac{2}{3}x\right|=\dfrac{5}{12}\)
\(\Leftrightarrow\left[{}\begin{matrix}\dfrac{4}{5}-\dfrac{2}{3}x=\dfrac{5}{12}\\\dfrac{4}{5}-\dfrac{2}{3}x=-\dfrac{5}{12}\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}\dfrac{2}{3}x=\dfrac{4}{5}-\dfrac{5}{12}\\\dfrac{2}{3}x=\dfrac{4}{5}-\left(-\dfrac{5}{12}\right)\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}\dfrac{2}{3}x=\dfrac{23}{60}\\\dfrac{2}{3}x=\dfrac{73}{60}\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{23}{60}:\dfrac{2}{3}\\x=\dfrac{73}{60}:\dfrac{2}{3}\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{23}{20}\\x=\dfrac{73}{20}\end{matrix}\right.\)
a)\(\dfrac{3}{7}x-2\dfrac{1}{3}=0,5\)
\(\dfrac{3}{7}x=0,5+2\dfrac{1}{3}\)
\(\dfrac{3}{7}x=\dfrac{17}{6}\)
\(x=\dfrac{17}{6}:\dfrac{3}{7}\)
\(x=\dfrac{119}{18}\)
b)\(\dfrac{4}{7}-\dfrac{2}{3}:x=1\dfrac{1}{4}\)
\(\dfrac{2}{3}:x=\dfrac{4}{7}-1\dfrac{1}{4}\)
\(x=\dfrac{2}{3}:\left(-\dfrac{19}{28}\right)\)
\(x=-\dfrac{56}{57}\)
c)\(\left(\dfrac{2}{3}x+2\dfrac{1}{4}\right):3\dfrac{1}{5}=0,75\)
\(\dfrac{2}{3}x+2\dfrac{1}{4}=0,75:3\dfrac{1}{5}\)
\(\dfrac{2}{3}x=\dfrac{15}{64}-2\dfrac{1}{4}\)
\(x=-\dfrac{129}{64}:\dfrac{2}{3}\)
\(x=-\dfrac{387}{128}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
a.\(\dfrac{-4}{5}-\left(\dfrac{2}{3}x+1\dfrac{1}{4}\right)=\dfrac{2}{7}\)
\(\left(\dfrac{2}{3}x+1\dfrac{1}{4}\right)=\dfrac{-4}{5}-\dfrac{2}{7}=\dfrac{-38}{35}\)
\(\dfrac{2}{3}x=\dfrac{-38}{35}-1\dfrac{1}{4}\)
\(\dfrac{2}{3}x=\dfrac{-327}{140}\Rightarrow x=\dfrac{-327}{140}:\dfrac{2}{3}=\dfrac{-981}{280}\)
Vậy \(x=\dfrac{-981}{280}\)
b. \(\dfrac{5}{6}+\left(\dfrac{3}{4}-\dfrac{1}{2}:x\right)=\dfrac{-2}{3}\)
\(\left(\dfrac{3}{4}-\dfrac{1}{2}:x\right)=\dfrac{-2}{3}-\dfrac{5}{6}=\dfrac{-3}{2}\)
\(\dfrac{1}{2}:x=\dfrac{3}{4}-\dfrac{-3}{2}\)
\(\dfrac{1}{2}:x=\dfrac{9}{4}\Rightarrow x=\dfrac{1}{2}:\dfrac{9}{4}=\dfrac{2}{9}\)
Vậy \(x=\dfrac{2}{9}\)
c. \(\left(\dfrac{4}{5}x-1\dfrac{1}{3}\right):\dfrac{3}{4}=0,7\)
\(\left(\dfrac{4}{5}x-1\dfrac{1}{3}\right)=0,7.\dfrac{3}{4}=\dfrac{21}{40}\)
\(\dfrac{4}{5}x=\dfrac{21}{40}+1\dfrac{1}{3}=\dfrac{223}{120}\)
\(\Rightarrow x=\dfrac{223}{120}:\dfrac{4}{5}=\dfrac{223}{96}\)
Vậy \(x=\dfrac{223}{96}\)
d. \(\dfrac{5}{6}-\dfrac{3}{4}x=1\dfrac{1}{3}+0,5x\)
\(0,5x+\dfrac{3}{4}x=\dfrac{5}{6}-1\dfrac{1}{3}\)
\(\dfrac{5}{4}x=\dfrac{-1}{2}\Rightarrow x=\dfrac{-1}{2}:\dfrac{5}{4}=\dfrac{-2}{5}\)
Vậy \(x=\dfrac{-2}{5}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
a)\(\dfrac{2}{3}x-\dfrac{5}{6}=1\dfrac{1}{4}\)
\(\dfrac{2}{3}x-\dfrac{5}{6}=\dfrac{5}{4}\)
\(\dfrac{2}{3}x=\dfrac{5}{4}+\dfrac{5}{6}\)
\(\dfrac{2}{3}x=\dfrac{25}{12}\)
\(x=\dfrac{25}{12}:\dfrac{2}{3}\)
=>\(x=\dfrac{25}{8}\)
a) \(\dfrac{2}{3}x-\dfrac{5}{6}=1\dfrac{1}{4}\) b) \(2\dfrac{1}{3}-\dfrac{4}{5}:x=0,2\)
\(\dfrac{2}{3}x-\dfrac{5}{6}=\dfrac{5}{4}\) \(\dfrac{7}{3}-\dfrac{4}{5}:x=\dfrac{1}{5}\)
\(\dfrac{2}{3}x=\dfrac{5}{4}-\dfrac{5}{6}\) \(\dfrac{4}{5}:x=\dfrac{7}{3}-\dfrac{1}{5}\)
\(\dfrac{2}{3}x=\dfrac{30}{24}-\dfrac{20}{24}\) \(\dfrac{4}{5}:x=\dfrac{35}{15}-\dfrac{3}{15}\)
\(\dfrac{2}{3}x=\dfrac{5}{12}\) \(\dfrac{4}{5}:x=\dfrac{32}{15}\)
\(x=\dfrac{5}{12}:\dfrac{2}{3}\) \(x=\dfrac{4}{5}:\dfrac{32}{15}\)
\(x=\dfrac{5}{12}:\dfrac{8}{12}\) \(x=\dfrac{4}{5}.\dfrac{15}{32}\)
\(x=\dfrac{5}{12}.\dfrac{12}{8}=\dfrac{5}{8}\) \(x=\dfrac{4.15}{5.32}\)
\(x=\dfrac{1.3}{1.8}=\dfrac{3}{8}\)
d)\(\left(\dfrac{4}{3}-\dfrac{1}{4}x\right)^3=\dfrac{-8}{27}\)
\(\left(\dfrac{4}{3}-\dfrac{1}{4}x\right)^3=\left(\dfrac{-2}{3}\right)^3\)
\(\Rightarrow\dfrac{4}{3}-\dfrac{1}{4}x=\dfrac{-2}{3}\)
\(\Rightarrow\dfrac{1}{4}x=\dfrac{4}{3}-\dfrac{-2}{3}\)
\(\Rightarrow\dfrac{1}{4}x=2\)
\(\Rightarrow x=2:\dfrac{1}{4}\)
\(\Rightarrow x=2.4=8\)
![](https://rs.olm.vn/images/avt/0.png?1311)
+)Đặt A= \(\dfrac{1}{99}+\dfrac{2}{98}+\dfrac{3}{97}+...+\dfrac{99}{1}\)
A= \(\dfrac{1}{99}+\dfrac{2}{98}+\dfrac{3}{97}+...+\left(1+1+1+...+1\right)\) (99 chữ số 1)
A= \(\left(\dfrac{1}{99}+1\right)+\left(\dfrac{2}{98}+1\right)+...+\left(\dfrac{98}{2}+1\right)+1\)
A= \(\dfrac{100}{99}+\dfrac{100}{98}+...+\dfrac{100}{2}+1\)
A= \(100.\left(\dfrac{1}{99}+\dfrac{1}{98}+...+\dfrac{1}{2}+\dfrac{1}{100}\right)\)
⇒ M= \(\dfrac{\dfrac{1}{99}+\dfrac{2}{98}+...+\dfrac{99}{1}}{\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{100}}\)
M= \(\dfrac{100.\left(\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{100}\right)}{\dfrac{1}{2}+\dfrac{1}{3}+.....+\dfrac{1}{100}}\)
M= 100 (1)
+) Đặt B= \(92-\dfrac{1}{9}-\dfrac{2}{10}-...-\dfrac{92}{100}\)
B= \(\left(1+1+1+...+1\right)-\dfrac{1}{9}-\dfrac{2}{10}-...-\dfrac{92}{100}\) ( 92 chữ số 1)
B= \(\left(1-\dfrac{1}{9}\right)+\left(1-\dfrac{2}{10}\right)+...+\left(1-\dfrac{92}{100}\right)\)
B= \(\dfrac{8}{9}+\dfrac{8}{10}+...+\dfrac{8}{100}\)
B= \(8.\left(\dfrac{1}{9}+\dfrac{1}{10}+...+\dfrac{1}{100}\right)\)
⇒ N= \(\dfrac{8.\left(\dfrac{1}{9}+\dfrac{1}{10}+...+\dfrac{1}{100}\right)}{\dfrac{1}{45}+\dfrac{1}{50}+\dfrac{1}{55}+...+\dfrac{1}{500}}\)
N= 8 (2)
Từ (1) và (2)⇒ \(\dfrac{M}{N}\) = \(\dfrac{100}{8}\)= \(\dfrac{25}{2}\)
Vậy \(\dfrac{M}{N}=\dfrac{25}{2}\)