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a. \(M=\left(\frac{12}{13}+\frac{21}{65}\right):\frac{81}{13}\)
\(M=\left(\frac{120}{130}+\frac{42}{130}\right).\frac{13}{81}\)
\(M=\frac{81}{65}.\frac{13}{81}\)
\(M=\frac{1}{5}\)
M=(12.107/13.107 + 21.10101/65.10101):27.3/13=(12/13 + 21/65):27.3/13
=[(12.5+21)/65]:27.3/13
=81/13.5 : 27.3/13 =81/(5.27.3)=1/5
Đs: M=1/5
a: \(A=\dfrac{7}{12}+\dfrac{5}{72}-\dfrac{11}{36}=\dfrac{42}{72}+\dfrac{5}{72}-\dfrac{22}{72}=\dfrac{25}{72}\)
b: \(B=\dfrac{8+5}{10}:\dfrac{-5}{13}=\dfrac{13}{10}\cdot\dfrac{13}{-5}=-\dfrac{169}{100}\)
c: \(C=\left(\dfrac{88}{132}-\dfrac{33}{132}+\dfrac{60}{132}\right):\left(\dfrac{55}{132}+\dfrac{132}{132}-\dfrac{84}{132}\right)\)
\(=\dfrac{88-33+60}{55+132-84}=\dfrac{115}{103}\)
= 18 x ( \(\dfrac{19}{21}\)+\(\dfrac{8}{9}\))+27 x \(\dfrac{17}{27}\)
=18 x \(\dfrac{113}{63}\)+27 x \(\dfrac{17}{27}\)
=\(\dfrac{226}{7}\)+17
=\(\dfrac{345}{7}\)
Chúc bạn học tốt!!
Giải:
1) (-8/13:3/7+-5/13:3/7).(-4)3.|-3|/7
=[7/3.(-8/13+-5/13)].-192/7
=[7/3.(-1)].-192/7
=-7/3.-192/7
=64
2) 75%-(5/2+5/3)+(-1/2)2
=3/4-25/6+1/4
=(3/4+1/4)-25/6
=1-25/6
=-19/6
Chúc bạn học tốt!
1) \(\left(\dfrac{-8}{13}:\dfrac{3}{7}+\dfrac{-5}{13}:\dfrac{3}{7}\right).\dfrac{\left(-4\right).|-3|}{7}\)
= \(\left[\left(\dfrac{-8}{13}+\dfrac{-5}{13}\right):\dfrac{3}{7}\right].\dfrac{-64.3}{7}\)
= \(\left[-1:\dfrac{3}{7}\right].\dfrac{-192}{7}\)
= \(\dfrac{-7}{3}.\dfrac{-192}{7}\)
= \(64\)
2) \(75\%-\left(\dfrac{5}{2}+\dfrac{5}{3}\right)+\left(-\dfrac{1}{2}\right)^2\)
= \(\dfrac{3}{4}-\dfrac{25}{6}+\dfrac{1}{4}\)
= \(\left(\dfrac{3}{4}+\dfrac{1}{4}\right)-\dfrac{25}{6}\)
= \(1-\dfrac{25}{6}\)
= \(\dfrac{-19}{6}\)
Chúc bạn học tốt !
a: \(=\dfrac{99}{100}:\left(\dfrac{3}{12}-\dfrac{1}{12}+\dfrac{4}{12}\right)-\dfrac{49}{25}\)
\(=\dfrac{99}{100}:\dfrac{1}{2}-\dfrac{49}{25}\)
\(=\dfrac{99}{50}-\dfrac{98}{50}=\dfrac{1}{50}\)
b: \(=\dfrac{13}{15}\cdot\dfrac{1}{4}\cdot3+\left(\dfrac{32}{60}-1-\dfrac{19}{60}\right):\dfrac{47}{24}\)
\(=\dfrac{39}{60}+\dfrac{-19}{60}\cdot\dfrac{24}{47}\)
=459/940
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Biến đổi thừa số tổng quát: \(1+\dfrac{1}{\left(k-1\right)\left(k+1\right)}\) \(=\dfrac{\left(k-1\right)\left(k+1\right)+1}{\left(k-1\right)\left(k+1\right)}\) \(=\dfrac{k^2}{\left(k-1\right)\left(k+1\right)}\).
Do đó \(1+\dfrac{1}{1.3}=\dfrac{2^2}{1.3}\), \(1+\dfrac{1}{2.4}=\dfrac{3^2}{2.4}\), \(1+\dfrac{1}{3.5}=\dfrac{4^2}{3.5}\),..., \(1+\dfrac{1}{2018.2020}=\dfrac{2019^2}{2018.2020}\), \(1+\dfrac{1}{2019.2021}=\dfrac{2020^2}{2019.2021}\). Từ đó suy ra \(\left(1+\dfrac{1}{1.3}\right)\left(1+\dfrac{1}{2.4}\right)...\left(1+\dfrac{1}{2019.2021}\right)\)
\(=\dfrac{2^2}{1.3}.\dfrac{3^2}{2.4}.\dfrac{4^2}{3.5}.\dfrac{5^2}{4.6}.\dfrac{6^2}{5.7}...\dfrac{2019^2}{2018.2020}.\dfrac{2020^2}{2019.2021}\)
\(=\dfrac{2.2020}{2021}=\dfrac{4040}{2021}\)
\(M=\left(\dfrac{12}{13}+\dfrac{21}{65}\right):\dfrac{9}{13}=\dfrac{81}{65}\cdot\dfrac{13}{9}=\dfrac{1}{5}\cdot9=\dfrac{9}{5}\)