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GL
1
27 tháng 6 2023
a: \(=\dfrac{2}{5}\cdot\dfrac{5}{6}\cdot\dfrac{3}{4}\cdot\dfrac{4}{3}=\dfrac{2}{6}=\dfrac{1}{3}\)
b: \(=\dfrac{123\cdot145+100}{123\cdot145+12\cdot145-45}=\dfrac{123\cdot145+100}{123\cdot145+1695}\)
=3587/3906
BA
2
TV
1
VD
21 tháng 3 2022
\(y-\dfrac{2}{3}=\dfrac{4}{5}\times\dfrac{3}{4}\\ y-\dfrac{2}{3}=\dfrac{3}{5}\\ y=\dfrac{3}{5}+\dfrac{2}{3}\\ y=\dfrac{9}{15}+\dfrac{10}{15}\\ y=\dfrac{19}{15}\)
NT
Nguyễn Thị Thương Hoài
Giáo viên
VIP
1 tháng 3 2023
\(\dfrac{7}{6}\) + \(\dfrac{5}{12}\) - \(\dfrac{1}{18}\) - 1
= \(\dfrac{42}{36}\) + \(\dfrac{15}{36}\) - \(\dfrac{2}{36}\) - \(\dfrac{36}{36}\)
= \(\dfrac{19}{36}\)
\(\dfrac{13}{6}\) + \(\dfrac{5}{18}+3-\dfrac{7}{12}\)
= \(\dfrac{78}{36}+\dfrac{10}{36}+\dfrac{108}{36}-\dfrac{21}{36}\)
= \(\dfrac{175}{36}\)
3 + \(\dfrac{11}{4}-\dfrac{1}{12}-\dfrac{3}{16}\)
= \(\dfrac{144}{48}+\dfrac{132}{48}-\dfrac{4}{48}-\dfrac{9}{48}\)
= \(\dfrac{263}{48}\)
\(\dfrac{2}{5}\) x \(\dfrac{7}{4}\) - \(\dfrac{2}{5}\) x \(\dfrac{3}{4}\)
= \(\dfrac{2}{5}\) x ( \(\dfrac{7}{4}\) - \(\dfrac{3}{4}\))
= \(\dfrac{2}{5}\) x \(\dfrac{4}{4}\)
= \(\dfrac{2}{5}\)
NH
0
F
23 tháng 1 2020
Bài giải
a, \(\frac{4}{5}-\frac{2}{3}+\frac{1}{5}-\frac{1}{3}\)
\(=\left(\frac{4}{5}+\frac{1}{5}\right)-\left(\frac{2}{3}+\frac{1}{3}\right)=1-1=0\)
b, \(\frac{2}{5}\text{ x }\frac{7}{4}-\frac{2}{5}\text{ x }\frac{3}{7}\)
\(=\frac{2}{5}\text{ x }\left(\frac{7}{4}-\frac{3}{7}\right)=\frac{2}{5}\text{ x }\frac{37}{28}=\frac{37}{70}\)
c, \(\frac{13}{4}\text{ x }\frac{2}{3}\text{ x }\frac{4}{13}\text{ x }\frac{3}{12}=\frac{13\text{ x }2\text{ x }4\text{ x }3}{4\text{ x }3\text{ x }13\text{ x }12}=\frac{1}{6}\)
d, \(\frac{75}{100}+\frac{18}{21}+\frac{19}{32}+\frac{1}{4}+\frac{3}{21}+\frac{13}{32}\)
\(=\frac{3}{4}+\frac{18}{21}+\frac{19}{32}+\frac{1}{4}+\frac{3}{21}+\frac{13}{32}\)
\(=\left(\frac{3}{4}+\frac{1}{4}\right)+\left(\frac{18}{21}+\frac{3}{21}\right)+\left(\frac{19}{32}+\frac{13}{32}\right)\)
\(=1+1+1\)
\(=3\)
e, \(\frac{2}{5}+\frac{6}{9}+\frac{3}{4}+\frac{3}{5}+\frac{1}{3}+\frac{1}{4}\)
\(=\frac{2}{5}+\frac{2}{3}+\frac{3}{4}+\frac{3}{5}+\frac{1}{3}+\frac{1}{4}\)
\(=\frac{1}{5}\left(2+3\right)+\frac{1}{3}\left(2+1\right)+\frac{1}{4}\left(3+1\right)\)
\(=\frac{1}{5}\cdot5+\frac{1}{3}\cdot3+\frac{1}{4}\cdot4\)
\(=1+1+1\)
\(=3\)
M
23 tháng 1 2020
a, \(\frac{4}{5}-\frac{2}{3}+\frac{1}{5}-\frac{1}{3}\)
\(=\left(\frac{4}{5}+\frac{1}{5}\right)-\left(\frac{2}{3}+\frac{1}{3}\right)=1-1=0\)
b, \(\frac{2}{5}\text{ x }\frac{7}{4}-\frac{2}{5}\text{ x }\frac{3}{7}\)
\(=\frac{2}{5}\text{ x }\left(\frac{7}{4}-\frac{3}{7}\right)=\frac{2}{5}\text{ x }\frac{37}{28}=\frac{37}{70}\)
c, \(\frac{13}{4}\text{ x }\frac{2}{3}\text{ x }\frac{4}{13}\text{ x }\frac{3}{12}=\frac{13\text{ x }2\text{ x }4\text{ x }3}{4\text{ x }3\text{ x }13\text{ x }12}=\frac{1}{6}\)
d, \(\frac{75}{100}+\frac{18}{21}+\frac{19}{32}+\frac{1}{4}+\frac{3}{21}+\frac{13}{32}\)
\(=\frac{3}{4}+\frac{18}{21}+\frac{19}{32}+\frac{1}{4}+\frac{3}{21}+\frac{13}{32}\)
\(=\left(\frac{3}{4}+\frac{1}{4}\right)+\left(\frac{18}{21}+\frac{3}{21}\right)+\left(\frac{19}{32}+\frac{13}{32}\right)\)
\(=1+1+1\)
\(=3\)
e, \(\frac{2}{5}+\frac{6}{9}+\frac{3}{4}+\frac{3}{5}+\frac{1}{3}+\frac{1}{4}\)
\(=\frac{2}{5}+\frac{2}{3}+\frac{3}{4}+\frac{3}{5}+\frac{1}{3}+\frac{1}{4}\)
\(=\frac{1}{5}\left(2+3\right)+\frac{1}{3}\left(2+1\right)+\frac{1}{4}\left(3+1\right)\)
\(=\frac{1}{5}\cdot5+\frac{1}{3}\cdot3+\frac{1}{4}\cdot4\)
\(=1+1+1\)
\(=3\)
ra5/6 nha
làm đầy đủ k nhé mấy đứa tiểu nawnh!