\(A=\dfrac{-5}{7}.\dfrac{8}{9}+\dfrac{-2}{7}.\dfrac{5}{9}+3\dfrac{4}{9}\)
K
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NT
Nguyễn Thị Thương Hoài
Giáo viên
VIP
5 tháng 8 2023
a, \(\dfrac{7}{8}\) \(\times\) \(\dfrac{3}{13}\) + \(\dfrac{4}{9}\) \(\times\) \(\dfrac{4}{13}\)
= \(\dfrac{1}{13}\) \(\times\)( \(\dfrac{21}{8}\) + \(\dfrac{16}{9}\))
= \(\dfrac{1}{13}\) \(\times\)( \(\dfrac{189}{72}\) + \(\dfrac{128}{72}\))
= \(\dfrac{1}{13}\) \(\times\) \(\dfrac{317}{73}\)
= \(\dfrac{317}{949}\)
b, \(\dfrac{6}{5}\) + \(\dfrac{7}{3}\) + \(\dfrac{8}{9}\)
= \(\dfrac{54}{45}\) + \(\dfrac{105}{45}\) + \(\dfrac{40}{45}\)
= \(\dfrac{199}{45}\)
c, 23 : \(\dfrac{5}{14}\) + \(\dfrac{6}{7}\) + \(\dfrac{4}{9}\)
= \(\dfrac{322}{5}\) + \(\dfrac{6}{7}\) + \(\dfrac{4}{9}\)
= \(\dfrac{20286}{315}\) + \(\dfrac{270}{315}\) + \(\dfrac{140}{315}\)
= \(\dfrac{20696}{315}\)
d, 4\(\dfrac{1}{4}\) + 7\(\dfrac{3}{7}\) - 2\(\dfrac{4}{17}\)
= 4 + \(\dfrac{1}{4}\) + 7 + \(\dfrac{3}{7}\) - 2 - \(\dfrac{4}{17}\)
= (4+7-2) + (\(\dfrac{1}{4}\) + \(\dfrac{3}{7}\) - \(\dfrac{4}{17}\))
= 9 + \(\dfrac{119}{476}\) + \(\dfrac{204}{476}\) - \(\dfrac{112}{476}\)
= 9\(\dfrac{211}{476}\) = \(\dfrac{4495}{476}\)
e, 8 - (9\(\dfrac{2}{11}\) + \(\dfrac{8}{33}\))
= 8 - 9 - \(\dfrac{2}{11}\) - \(\dfrac{8}{33}\)
= -1 - \(\dfrac{2}{11}\) - \(\dfrac{8}{33}\)
= \(\dfrac{-33}{33}\) - \(\dfrac{-6}{33}\) - \(\dfrac{8}{33}\)
= - \(\dfrac{47}{33}\)
C
15 tháng 5 2022
8/9 x 5/4 + 3/5 = 10/9 + 3/5 = 77/45
5/8 + 1/4 = 5/9 + 2/8 = 7/8
1/2 + 5/9 = 9/18 + 5/9 = 14/9
9/8 x 3/4 - 4/7 = 27/32 - 4/7 = 61/224
O
15 tháng 5 2022
a) \(\dfrac{8}{9}:\dfrac{4}{5}+\dfrac{3}{5}=\dfrac{8}{9}\times\dfrac{5}{4}+\dfrac{2}{5}=\dfrac{10}{9}+\dfrac{2}{5}=\dfrac{50}{45}+\dfrac{18}{45}=\dfrac{68}{45}\)
b) \(\dfrac{5}{8}+\dfrac{1}{4}=\dfrac{5}{8}+\dfrac{2}{8}=\dfrac{6}{8}=\dfrac{3}{4}\)
c) \(\dfrac{7}{8}\times\dfrac{4}{7}+\dfrac{5}{9}=\dfrac{1}{2}+\dfrac{5}{9}=\dfrac{9}{18}+\dfrac{10}{18}=\dfrac{19}{18}\)
d) \(\dfrac{9}{8}:\dfrac{4}{3}-\dfrac{4}{7}=\dfrac{9}{8}\times\dfrac{3}{4}-\dfrac{4}{7}=\dfrac{27}{32}-\dfrac{4}{7}=\dfrac{61}{244}\)
15 tháng 3 2022
\(a,\dfrac{3}{5}+\dfrac{-5}{9}=\dfrac{27-25}{45}=\dfrac{2}{49}.\)
\(c,\dfrac{-27}{23}+\dfrac{5}{21}+\dfrac{4}{23}+\dfrac{16}{21}+\dfrac{1}{2}=\dfrac{-23}{23}+\dfrac{21}{21}+\dfrac{1}{2}=-1+1+\dfrac{1}{2}=\dfrac{1}{2}.\)
\(d,\dfrac{-8}{9}+\dfrac{1}{9}.\dfrac{2}{9}+\dfrac{1}{9}.\dfrac{7}{9}=\dfrac{-8}{9}+\dfrac{1}{9}.\left(\dfrac{2}{9}+\dfrac{7}{9}\right)=\dfrac{-8}{9}+\dfrac{1}{9}.1=\dfrac{-8+1}{9}=\dfrac{-7}{9}.\)
31 tháng 5 2021
a) \(A=\dfrac{7}{8}\left(-\dfrac{5}{9}-\dfrac{4}{9}\right)+5\dfrac{7}{8}\)
\(A=\dfrac{7}{8}.\left(-1\right)+5\dfrac{7}{8}=5\dfrac{7}{8}-\dfrac{7}{8}=5\).
31 tháng 5 2021
\(B=\dfrac{1}{4}.\dfrac{8}{5}.\dfrac{25}{16}.\dfrac{-7}{4}=\dfrac{-35}{32}\)
O
2 tháng 5 2022
a) \(\dfrac{3}{7}+4=\dfrac{3}{7}+\dfrac{4}{1}=\dfrac{3}{7}+\dfrac{28}{7}=\dfrac{31}{7}\)
b) \(\dfrac{5}{9}\times3=\dfrac{5\times3}{9}=\dfrac{15}{9}=\dfrac{5}{3}\)
c) \(\dfrac{7}{8}-\dfrac{2}{9}=\dfrac{63}{72}-\dfrac{16}{72}=\dfrac{47}{72}\)
d) \(5-\dfrac{3}{4}=\dfrac{5}{1}-\dfrac{3}{4}=\dfrac{20}{4}-\dfrac{3}{4}=\dfrac{17}{4}\)
e) \(\dfrac{5}{7}:6=\dfrac{5}{7}:\dfrac{6}{1}=\dfrac{5}{7}\times\dfrac{1}{6}=\dfrac{5}{42}\)
f) \(\dfrac{5}{6}\times\dfrac{2}{7}=\dfrac{5\times2}{6\times7}=\dfrac{10}{42}=\dfrac{5}{21}\)
g) \(3+\dfrac{3}{4}=\dfrac{3}{1}+\dfrac{3}{4}=\dfrac{12}{4}+\dfrac{3}{4}=\dfrac{15}{4}\)
h) \(\dfrac{3}{5}\times\dfrac{4}{8}=\dfrac{3\times4}{5\times8}=\dfrac{12}{40}=\dfrac{3}{10}\)
i) \(5:\dfrac{3}{8}=\dfrac{5}{1}:\dfrac{3}{8}=\dfrac{5}{1}\times\dfrac{8}{3}=\dfrac{40}{3}\)
2 tháng 5 2022
\(\dfrac{3}{7}+\dfrac{4}{1}=\dfrac{3}{7}+\dfrac{28}{7}=\dfrac{31}{7}\)
\(\dfrac{5}{9}\times3=\dfrac{5}{9}\times\dfrac{3}{1}=\dfrac{15}{9}=\dfrac{5}{3}\)
\(\dfrac{7}{8}-\dfrac{2}{9}=\dfrac{63}{72}-\dfrac{16}{72}=\dfrac{47}{72}\)
\(\dfrac{5}{1}-\dfrac{3}{4}=\dfrac{20}{4}-\dfrac{3}{4}=\dfrac{17}{4}\)
\(\dfrac{5}{7}:\dfrac{6}{1}=\dfrac{5}{7}\times\dfrac{1}{6}=\dfrac{5}{42}\)
\(\dfrac{5}{6}\times\dfrac{2}{7}=\dfrac{5\times2}{6\times7}=\dfrac{10}{42}=\dfrac{5}{21}\)
\(\dfrac{3}{1}+\dfrac{3}{4}=\dfrac{12}{4}+\dfrac{3}{4}=\dfrac{15}{4}\)
\(\dfrac{3}{5}\times\dfrac{4}{8}=\dfrac{3}{5}\times\dfrac{1}{2}=\dfrac{3\times1}{5\times2}=\dfrac{3}{10}\)
\(\dfrac{5}{1}:\dfrac{3}{8}=\dfrac{5}{1}\times\dfrac{8}{3}=\dfrac{40}{3}\)
16 tháng 4 2022
8: \(=\dfrac{-5}{9}-\dfrac{4}{9}+\dfrac{8}{15}+\dfrac{7}{15}-\dfrac{2}{11}=\dfrac{-2}{11}\)
9: =2/7-2/5+5/7=1-2/5=3/5
10: \(=\dfrac{7}{19}\left(\dfrac{8}{11}+\dfrac{3}{11}\right)-\dfrac{12}{19}=\dfrac{-5}{19}\)
11: \(=\dfrac{-5}{7}\left(\dfrac{2}{11}+\dfrac{9}{11}\right)=\dfrac{-5}{7}\)
29 tháng 10 2023
a: \(\dfrac{3}{8}+\dfrac{7}{8}=\dfrac{3+7}{8}=\dfrac{10}{8}=\dfrac{5}{4}\)
b: \(\dfrac{7}{9}-\dfrac{4}{9}=\dfrac{7-4}{9}=\dfrac{3}{9}=\dfrac{1}{3}\)
c: \(\dfrac{5}{6}+\dfrac{1}{8}=\dfrac{20}{24}+\dfrac{3}{24}=\dfrac{20+3}{24}=\dfrac{23}{24}\)
d: \(\dfrac{9}{15}-\dfrac{2}{5}=\dfrac{3}{5}-\dfrac{2}{5}=\dfrac{3-2}{5}=\dfrac{1}{5}\)
e: \(\dfrac{2}{5}+\dfrac{3}{5}+\dfrac{9}{5}=\dfrac{2+3+9}{5}=\dfrac{14}{5}\)
g: \(\dfrac{8}{10}-\dfrac{1}{10}-\dfrac{3}{10}=\dfrac{8-1-3}{10}=\dfrac{4}{10}=\dfrac{2}{5}\)
h: \(\dfrac{23}{7}-\dfrac{4}{7}+\dfrac{2}{7}=\dfrac{23-4+2}{7}=\dfrac{21}{7}=3\)
KN
8 tháng 5 2022
a. 7/8 x 2/5 =7/20
b.9/4-5/6=17/12
c.4/7x2/5=8/35
d.3/5+2=13/5
e. 4-3/5=17/5
g.3x4/9=4/3
h.9/5:2=9/10
C
8 tháng 5 2022
a) \(\dfrac{14}{40}=\dfrac{7}{20}\)
b) \(\dfrac{27}{12}-\dfrac{10}{12}=\dfrac{17}{12}\)
c) \(\dfrac{8}{35}\)
d) \(\dfrac{3}{5}+\dfrac{10}{5}=\dfrac{13}{5}\)
e) \(\dfrac{20}{5}-\dfrac{3}{5}=\dfrac{17}{5}\)
g) \(\dfrac{12}{9}=\dfrac{4}{3}\)
h) \(\dfrac{9}{5}x\dfrac{1}{2}=\dfrac{9}{10}\)
HQ
Hà Quang Minh
Giáo viên
27 tháng 11 2023
a) $\frac{5}{9}:\frac{7}{3} = \frac{5}{9} \times \frac{3}{7} = \frac{{15}}{{63}} = \frac{5}{{21}}$
b) $\frac{2}{7}:\frac{8}{5} = \frac{2}{7} \times \frac{5}{8} = \frac{{10}}{{56}} = \frac{5}{{28}}$
c) $\frac{4}{{15}}:\frac{2}{5} = \frac{4}{{15}} \times \frac{5}{2} = \frac{{20}}{{30}} = \frac{2}{3}$
d) $\frac{7}{9}:\frac{{21}}{9} = \frac{7}{9} \times \frac{9}{{21}} = \frac{{63}}{{189}} = \frac{1}{3}$
A=\(\dfrac{-5}{7}.\dfrac{8}{9}+\dfrac{-2}{7}.\dfrac{5}{9}+\dfrac{31}{9}\)
=\(\dfrac{-40}{63}+\dfrac{-10}{63}+\dfrac{217}{63}\)
=\(\dfrac{-40+\left(-10\right)+217}{63}\)
=\(\dfrac{167}{63}\)