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PT
2
23 tháng 8 2016
\(\left(-2\right).\frac{-38}{21}.\frac{-7}{4}.\frac{-3}{8}\)
\(=\left(-2.\frac{-7}{4}.\frac{-3}{8}\right).\frac{-38}{21}\)
\(=\left(\frac{7}{2}.\frac{-3}{8}\right).\frac{-38}{21}\)
\(=\frac{-21}{16}.\frac{-38}{21}\)
\(=\frac{-38}{-16}=\frac{19}{8}\)
23 tháng 8 2016
\(\left(-2\right).\frac{-38}{21}.\frac{-7}{4}.\left(\frac{-3}{8}\right)\)
\(=\left(-2.\frac{-7}{2}.\frac{-3}{8}\right).\frac{-38}{21}\)
\(=\left(7.\frac{-3}{8}\right).\frac{-38}{21}\)
\(=\frac{-21}{8}.\frac{-38}{21}\)
\(=\frac{-38}{-8}=\frac{19}{4}\)
HQ
Hà Quang Minh
Giáo viên
19 tháng 9 2023
a)
\(\begin{array}{l}0,75 - \frac{5}{6} + 1\frac{1}{2} = \frac{3}{4} - \frac{5}{6} + \frac{3}{2}\\ = \frac{9}{{12}} - \frac{{10}}{{12}} + \frac{{18}}{{12}} = \frac{{17}}{{12}}\end{array}\)
b)
\(\begin{array}{l}\frac{3}{7} + \frac{4}{{15}} + \left( {\frac{{ - 8}}{{21}}} \right) + \left( { - 0,4} \right) = \frac{3}{7} + \frac{4}{{15}} - \frac{8}{{21}} - \frac{2}{5}\\ = \left( {\frac{3}{7} - \frac{8}{{21}}} \right) + \left( {\frac{4}{{15}} - \frac{2}{5}} \right)\\ = \left( {\frac{9}{{21}} - \frac{8}{{21}}} \right) + \left( {\frac{4}{{15}} - \frac{6}{{15}}} \right)\\ = \frac{1}{{21}} + \left( {\frac{{ - 2}}{{15}}} \right)\\ = \frac{5}{{105}} - \frac{{14}}{{105}}\\ = \frac{{ - 9}}{{105}} = \frac{{ - 3}}{{35}}\end{array}\)
c)
\(\begin{array}{l}0,625 + \left( {\frac{{ - 2}}{7}} \right) + \frac{3}{8} + \left( {\frac{{ - 5}}{7}} \right) + 1\frac{2}{3}\\ = \frac{5}{8} + \left( {\frac{{ - 2}}{7}} \right) + \frac{3}{8} - \frac{5}{7} + \frac{5}{3}\\ = \left( {\frac{5}{8} + \frac{3}{8}} \right) + \left( {\frac{{ - 2}}{7} - \frac{5}{7}} \right) + \frac{5}{3}\\ = 1 - 1 + \frac{5}{3} = \frac{5}{3}\end{array}\)
d)
\(\begin{array}{l}\left( { - 3} \right).\left( {\frac{{ - 38}}{{21}}} \right).\left( {\frac{{ - 7}}{6}} \right).\left( { - \frac{3}{{19}}} \right)\\ = \frac{{ - 3.\left( { - 38} \right).\left( { - 7} \right).\left( { - 3} \right)}}{{21.6.19}}\\ = \frac{{3.38.7.3}}{{21.6.19}}\\ = \frac{{3.2.19.7.3}}{{3.7.3.2.19}}\\ = 1\end{array}\)
e)
\(\begin{array}{l}\left( {\frac{{11}}{{18}}:\frac{{22}}{9}} \right).\frac{8}{5} = \left( {\frac{{11}}{{18}}.\frac{9}{{22}}} \right).\frac{8}{5}\\ = \frac{{11.9.4.2}}{{9.2.2.11.5}} = \frac{2}{5}\end{array}\)
g)
\(\left[ {\left( {\frac{{ - 4}}{5}} \right).\frac{5}{8}} \right]:\left( {\frac{{ - 25}}{{12}}} \right) = \frac{{ - 20}}{{40}}:\left( {\frac{{ - 25}}{{12}}} \right)\\ = \frac{{ - 1}}{2}.\frac{{ - 12}}{{25}} = \frac{6}{{25}}\)
17 tháng 6 2019
\(\left(-2\right).\left(\frac{-38}{21}\right).\left(\frac{-7}{4}\right).\left(\frac{3}{-8}\right)\)
\(=\frac{\left(-2\right).\left(-38\right).\left(-7\right).\left(-3\right)}{21.4.8}\)
\(=\frac{19}{8}\)
HA
17 tháng 6 2019
\(\left(-2\right)\cdot\frac{-38}{21}\cdot\frac{-7}{4}\cdot\left(\frac{-3}{8}\right)\)
\(=\frac{-2\cdot\left(-38\right)\cdot\left(-7\right)\cdot\left(-3\right)}{21\cdot4\cdot8}=\frac{-2\cdot\left(-38\right)\cdot21}{21\cdot4\cdot8}=\frac{-2\cdot\left(-38\right)}{32}=\frac{19}{8}\)
PK
1 tháng 9 2019
a, \(\frac{1}{5.6}+\frac{1}{6.7}+...+\frac{1}{x\left(x+1\right)}=\frac{13}{90}\)
⇒ \(\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}+...+\frac{1}{x}-\frac{1}{x+1}=\frac{13}{90}\)
⇒ \(\frac{1}{5}-\frac{1}{x+1}=\frac{13}{90}\)
⇒ \(\frac{1}{x+1}=\frac{1}{5}-\frac{13}{90}\)
⇒ \(\frac{1}{x+1}=\frac{18}{90}-\frac{13}{90}\)
⇒ \(\frac{1}{x+1}=\frac{1}{18}\)
⇒ x + 1 = 18
⇒ x = 17
Vậy x = 17
b, \(\frac{1}{1.4}+\frac{1}{4.7}+\frac{1}{7.10}+...+\frac{1}{x\left(x+3\right)}=\frac{49}{148}\)
⇒ \(\frac{3}{1.4}+\frac{3}{4.7}+\frac{3}{7.10}+...+\frac{3}{x\left(x+3\right)}=\frac{49.3}{148}\)
⇒ \(1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+...+\frac{1}{x}-\frac{1}{x+3}=\frac{147}{148}\)
⇒ \(1-\frac{1}{x+3}=\frac{147}{148}\)
⇒ \(\frac{1}{x+3}=1-\frac{147}{148}\)
⇒ \(\frac{1}{x+3}=\frac{1}{148}\)
⇒ x + 3 = 148
⇒ x = 145
Vậy x = 145
(-2).-38/21.-7/4.-3/8
=19/8
bài nãy dễ mà bạn
\(\left(-2\right).\frac{-38}{21}.\frac{-7}{4}.\frac{\left(-3\right)}{8}=\frac{-2.\left(-38\right).\left(-7\right).\left(-3\right)}{21.4.8}=\frac{4.19.21}{21.4.8}=\frac{19}{8}\)