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NT
Nguyễn Thị Thương Hoài
Giáo viên
VIP
27 tháng 3 2024
\(\left(\dfrac{5}{7}-\dfrac{7}{7}\right)-\left[0,2-\left(-\dfrac{2}{7}-\dfrac{1}{10}\right)\right]\)
=\(-\dfrac{2}{7}-\left[\dfrac{1}{5}+\dfrac{2}{7}+\dfrac{1}{10}\right]\)
=\(-\dfrac{2}{7}-\dfrac{1}{5}-\dfrac{2}{7}-\dfrac{1}{10}\)
=\(\left(-\dfrac{2}{7}-\dfrac{2}{7}\right)-\left(\dfrac{1}{5}+\dfrac{1}{10}\right)\)
=\(-\dfrac{4}{7}-\left(\dfrac{2}{10}+\dfrac{1}{10}\right)\)
=\(-\dfrac{4}{7}-\dfrac{3}{10}\)
=\(-\dfrac{40}{70}-\dfrac{21}{70}\)
=\(-\dfrac{61}{70}\)
NT
Nguyễn Thị Thương Hoài
Giáo viên
VIP
27 tháng 3 2024
(3 - \(\dfrac{1}{4}\) + \(\dfrac{2}{3}\)) - (5 - \(\dfrac{1}{3}\) - \(\dfrac{5}{6}\)) - (6 - \(\dfrac{7}{4}\) - \(\dfrac{3}{2}\))
= 3 - \(\dfrac{1}{4}\) + \(\dfrac{2}{3}\) - 5 + \(\dfrac{1}{3}\) + \(\dfrac{5}{6}\) - 6 + \(\dfrac{7}{4}\) + \(\dfrac{3}{2}\)
= (3 - 5 - 6) + ( \(\dfrac{7}{4}\) - \(\dfrac{1}{4}\)) + (\(\dfrac{2}{3}\) + \(\dfrac{1}{3}\)) + \(\dfrac{3}{2}\) + \(\dfrac{5}{6}\)
= - 8 + \(\dfrac{3}{2}\) + 1 + \(\dfrac{3}{2}\) + \(\dfrac{5}{6}\)
= (- 8 + 1) + (\(\dfrac{3}{2}\) + \(\dfrac{3}{2}\)) + \(\dfrac{5}{6}\)
= -7 + 3 + \(\dfrac{5}{6}\)
= - 4 + \(\dfrac{5}{6}\)
= \(\dfrac{-19}{6}\)
PN
14 tháng 10 2017
a) \(1\dfrac{7}{20}:2,7+2,7:1,35+\left(0,4:2\dfrac{1}{2}\right).\left(4,2-1\dfrac{3}{40}\right)\)
\(=\dfrac{27}{20}:\dfrac{27}{10}+\dfrac{27}{10}:\dfrac{27}{10}+\left(\dfrac{2}{5}:\dfrac{5}{2}\right).\left(\dfrac{21}{5}-\dfrac{43}{40}\right)\)
\(=1+1+\dfrac{4}{25}.\dfrac{25}{8}\)
\(=2+\dfrac{1}{2}\)
\(=2\dfrac{1}{2}\)
PN
14 tháng 10 2017
b) \(\left(6\dfrac{3}{5}-3\dfrac{3}{14}\right).5\dfrac{5}{6}:\left(21-1.25\right):2,5\)
\(=\left(\dfrac{33}{5}-\dfrac{45}{14}\right).\dfrac{35}{6}:\left(-4\right):2,5\)
\(=\left(\dfrac{462}{60}-\dfrac{225}{60}\right).\dfrac{35}{6}.\dfrac{1}{-4}:\dfrac{5}{2}\)
\(=\dfrac{237}{60}.\dfrac{35}{6}.\dfrac{1}{-4}.\dfrac{2}{5}\)
\(=\dfrac{3.79.7.5.2}{5.14.3.2.\left(-4\right).5}\)
\(=\dfrac{79.7}{14.\left(-4\right).5}=\dfrac{553}{-280}\) (số xấu :v)
KH
28 tháng 2 2018
a) \(x=\dfrac{1}{4}+\dfrac{2}{13}\)
\(x=\dfrac{13}{52}+\dfrac{8}{52}\)
⇒ \(x=\dfrac{21}{52}\)
b) \(\dfrac{x}{3}=\dfrac{2}{3}+\dfrac{-1}{7}\)
\(\dfrac{x}{3}=\dfrac{14}{21}+\dfrac{-3}{21}\)
\(\dfrac{x}{3}=\dfrac{11}{21}\)
⇒ \(x=\dfrac{11.3}{21}=\dfrac{33}{21}\)
⇒ \(x=\dfrac{11}{7}\)
c) \(\dfrac{-8}{3}+\dfrac{1}{3}< x< \dfrac{-2}{7}+\dfrac{-5}{7}\)
\(\dfrac{-17}{7}< x< -1\)
⇒ \(-17< x< -7\)
⇒ \(x\in\left\{-16;-15,....;-6\right\}\)
d) \(\dfrac{1}{6}+\dfrac{2}{5}\)
\(=\dfrac{5}{30}+\dfrac{12}{30}\)
\(=\dfrac{17}{30}\)
e) \(\dfrac{3}{5}+\dfrac{-7}{4}\)
\(=\dfrac{12}{20}+\dfrac{-35}{20}\)
\(=\dfrac{-23}{20}\)
f) \(\dfrac{4}{13}+\dfrac{-12}{30}\)
\(=\dfrac{4}{13}+\dfrac{-2}{5}\)
\(=\dfrac{20}{65}+\dfrac{-26}{65}\)
\(=\dfrac{-6}{65}\)
g) \(\dfrac{-3}{29}+\dfrac{16}{58}\)
\(=\dfrac{-6}{58}+\dfrac{16}{58}\)
\(=\dfrac{10}{58}\)
h) \(\dfrac{8}{40}+\dfrac{-36}{45}\)
\(=\dfrac{1}{5}+\dfrac{-4}{5}\)
\(=\dfrac{-3}{5}\)
j) \(\dfrac{-8}{18}+\dfrac{15}{27}\)
\(=\dfrac{-2}{9}+\dfrac{5}{9}\)
\(=\dfrac{3}{9}\)
\(=\dfrac{1}{3}\)
Bài làm:
\(7\left|\frac{3}{5}x-40\%\right|-\frac{3}{5}:\frac{1}{2}=0,2\)
\(\Leftrightarrow7\left|\frac{3}{5}x-\frac{2}{5}\right|-\frac{3}{5}.2=\frac{1}{5}\)
\(\Leftrightarrow7\left|\frac{3}{5}x-\frac{2}{5}\right|=\frac{1}{5}+\frac{6}{5}\)
\(\Leftrightarrow\frac{7}{5}\left|3x-2\right|=\frac{7}{5}\)
\(\Leftrightarrow\left|3x-2\right|=1\)
\(\Leftrightarrow\orbr{\begin{cases}3x-2=1\\3x-2=-1\end{cases}\Leftrightarrow\orbr{\begin{cases}3x=3\\3x=1\end{cases}\Leftrightarrow}\orbr{\begin{cases}x=1\\x=\frac{1}{3}\end{cases}}}\)
Vậy \(x=1\)hoặc \(x=\frac{1}{3}\)
Học tốt!!!!