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\(\frac{\frac{4}{3}+\frac{4}{7}-\frac{2}{14}}{-1-\frac{3}{7}+\frac{3}{28}}\)
K
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NP
2
KN
14 tháng 4 2019
\(\frac{\frac{2}{3}+\frac{2}{7}-\frac{1}{14}}{-1-\frac{3}{7}+\frac{3}{28}}\)
\(=\frac{37}{\frac{42}{-\frac{37}{28}}}\)
\(=-\frac{2}{3}\)
14 tháng 4 2019
=28/42+12/42-3/42/-28/28-12/28+3/28=40/42-3/42/-40/28+3/28=37/42/-37/42=-2/3
NT
Nguyễn Thị Thương Hoài
Giáo viên
VIP
6 tháng 7
Bài 1:
A = \(\frac15\) + \(\frac{3}{17}\) - \(\frac43\) + (\(\frac45\) - \(\frac{3}{17}\) + \(\frac13\)) - \(\frac17\) + (- \(\frac{14}{30}\))
A = \(\frac15\) + \(\frac{3}{17}\) - \(\frac43\) + \(\frac45\) - \(\frac{3}{17}\) + \(\frac13\) - \(\frac17\) - \(\frac{14}{30}\)
A = (\(\frac15\) + \(\frac45\)) + (\(\frac{3}{17}\) - \(\frac{3}{17}\)) - (\(\frac43-\frac13\)) - \(\frac{30}{210}\) - \(\frac{98}{210}\)
A = 1 + 0 - 1 - (\(\frac{30}{210}+\frac{98}{210}\))
A = 1 - 1 - \(\frac{228}{210}\)
A = 0 - \(\frac{128}{210}\)
A = - \(\frac{64}{105}\)
NT
Nguyễn Thị Thương Hoài
Giáo viên
VIP
6 tháng 7
Bài 2:
B= (\(\frac58\) - \(\frac{4}{12}\) + \(\frac32\)) - (\(\frac58\) + \(\frac{9}{13}\)) - (\(\frac{-3}{2}\)) + \(\frac{7}{-15}\)
B = \(\frac58\) - \(\frac{4}{12}\) + \(\frac32\) - \(\frac58\) - \(\frac{9}{13}\) + \(\frac32\) - \(\frac{7}{15}\)
B = (\(\frac58\) - \(\frac58\)) + (\(\frac32\) + \(\frac32\)) - (\(\frac13\) + \(\frac{9}{13}\) + \(\frac{7}{15}\))
B = 0 + 3 - (\(\frac{65}{195}\) + \(\frac{135}{195}\) + \(\frac{91}{195}\))
B = 3 - (\(\frac{200}{195}\) + \(\frac{91}{195}\))
B = 3 - \(\frac{97}{65}\)
B = \(\frac{195}{65}\) - \(\frac{97}{65}\)
B = \(\frac{98}{65}\)
B
24 tháng 3 2019
a) \(\frac{14}{21}+1-\left|\frac{1}{3}-1\right|\)
\(=\frac{2}{3}+1-\frac{2}{3}\)
\(=1+\left(\frac{2}{3}-\frac{1}{3}\right)\)
\(=1\)
b) \(\frac{1}{3}-\left|\frac{-1}{4}+\frac{5}{6}\right|-\left|\frac{-7}{12}\right|\)
\(=\frac{1}{3}-\frac{7}{12}-\frac{7}{12}\)
\(=-\frac{5}{6}\)
N
20 tháng 6 2017
\(\frac{1+\frac{1}{7}+\frac{1}{7^2}-\frac{1}{7^3}}{4+\frac{4}{7}+\frac{4}{7^2}-\frac{4}{7^3}}.\frac{858585}{313131}.\left(-1\frac{14}{17}\right)\)\
\(=\frac{1+\frac{1}{7}+\frac{1}{7^2}-\frac{1}{7^3}}{4.\left(1+\frac{1}{7}+\frac{1}{7^2}-\frac{1}{7^3}\right)}.\frac{85}{31}.\left(-\frac{31}{17}\right)\)
\(=\frac{1}{4}.\frac{85}{31}.\left(-\frac{31}{17}\right)\)
\(=-\frac{5}{4}\)
20 tháng 6 2017
\(=\frac{1+\frac{1}{7}+\frac{1}{7^2}-\frac{1}{7^3}}{4\left(1+\frac{1}{7}+\frac{1}{7^2}-\frac{1}{7^3}\right)}.\frac{85.10101}{31.10101}.\left(-\frac{31}{17}\right).\)
\(=\frac{1}{4}.\frac{3.17}{31}.\left(-\frac{31}{17}\right)=-\frac{3}{4}\)
TM
21 tháng 9 2017
\(P=\frac{\frac{3}{7}-\frac{3}{13}+\frac{3}{37}}{\frac{5}{7}-\frac{5}{17}+\frac{5}{37}}+\frac{\frac{1}{2}-\frac{1}{3}+\frac{1}{4}-\frac{1}{5}}{\frac{7}{5}-\frac{7}{4}+\frac{7}{3}-\frac{7}{2}}\)
\(=\frac{3\left(\frac{1}{7}-\frac{1}{13}+\frac{1}{37}\right)}{5\left(\frac{1}{7}-\frac{1}{17}+\frac{1}{37}\right)}+\frac{\frac{1}{2}-\frac{1}{3}+\frac{1}{4}-\frac{1}{5}}{-7\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{4}-\frac{1}{5}\right)}\)
\(=\frac{3}{5}+\frac{1}{-7}\)
\(=\frac{16}{35}\)
\(\frac{\frac{4}{3}+\frac{4}{7}-\frac{2}{14}}{-1-\frac{3}{7}+\frac{3}{28}}=\frac{4.\left(\frac{1}{3}+\frac{1}{7}-\frac{1}{28}\right)}{-3.\left(\frac{1}{3}+\frac{1}{7}-\frac{1}{28}\right)}=\frac{-4}{3}\)