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NT
Nguyễn Thị Thương Hoài
Giáo viên
VIP
25 tháng 5 2022
A = \(\dfrac{-5}{12}\)x\(\dfrac{2}{11}\)+\(\dfrac{-5}{12}\)x\(\dfrac{9}{11}\)+\(\dfrac{5}{12}\)
A= \(\dfrac{-5}{12}\)(\(\dfrac{2}{11}\)+\(\dfrac{9}{11}\)-1)
A = 0
DT
3
HN
30 tháng 4 2017
\(\dfrac{-5}{7}.\dfrac{2}{11}-\dfrac{5}{11}.\dfrac{9}{7}-2\dfrac{5}{7}\)
\(=\dfrac{-2}{7}.\dfrac{5}{11}-\dfrac{5}{11}.\dfrac{9}{7}-\dfrac{19}{7}\)
\(=\dfrac{5}{11}.\left(\dfrac{-2}{7}-\dfrac{9}{7}\right)-\dfrac{19}{7}\)
\(=\dfrac{5}{11}.\dfrac{-11}{7}-\dfrac{19}{7}\)
\(=\dfrac{-5}{7}-\dfrac{19}{7}\)
\(=\dfrac{-24}{7}\)
30 tháng 4 2017
\(\dfrac{-5}{7}.\dfrac{2}{11}-\dfrac{5}{11}.\dfrac{9}{7}-2\dfrac{5}{7}\)
\(=\dfrac{-2}{7}.\dfrac{5}{11}-\dfrac{5}{11}.\dfrac{9}{7}-\dfrac{19}{7}\)
\(=\dfrac{5}{7}\left(\dfrac{-2}{7}-\dfrac{9}{7}\right)-\dfrac{19}{7}\)
\(=\dfrac{5}{7}.\dfrac{-11}{7}-\dfrac{19}{7}\)
\(=\dfrac{5.\left(-11\right)}{11.7}-\dfrac{19}{7}\)
\(=\dfrac{-5.11}{11.7}-\dfrac{19}{7}\)
\(=\dfrac{-5}{7}-\dfrac{19}{7}\)
\(=\dfrac{-5-19}{7}\)
\(=\dfrac{\left(-5\right)+\left(-19\right)}{7}\)
\(=\dfrac{-24}{7}\)
DT
1
12 tháng 10 2017
\(\frac{5^{11}.7^{12}+5^{11}.7^{11}}{5^{12}.7^{12}+9.5^{11}.7^{11}}\)
=\(\frac{5^{11}.7^{11}\left(7+1\right)}{5^{11}.7^{11}\left(5.7+9\right)}\)
= \(\frac{8}{35+9}\)
= \(\frac{8}{44}\)
= \(\frac{2}{11}\)
D
0
I
0
OM
5
PH
1
22 tháng 3 2016
\(\frac{5^{11}\cdot7^{12}+5^{11}\cdot7^{11}}{5^{12}\cdot7^{11}+9\cdot5^{11}\cdot7^{11}}\)
=\(\frac{5^{11}\cdot\left(7^{12}+7^{11}\right)}{5^{11}\left(5\cdot7^{11}+9\cdot7^{11}\right)}\)
=\(\frac{7^{12}+7^{11}}{5\cdot7^{11}+9\cdot7^{11}}\)
=\(\frac{7^{11}\left(7+1\right)}{7^{11}\left(5+9\right)}\)
=\(\frac{8}{14}=\frac{4}{7}\)
Kết quả này chắc chắn là đúng đấy
6 tháng 3 2020
a)\(\frac{3^{10}.\left(-5\right)^{21}}{\left(-5\right)^{20}.3^{12}}=\frac{-5}{9}\)
b)\(\frac{\left(-11\right)^5.13^7}{11^5.13^8}=-\frac{1}{13}\)
c)\(\frac{2^{10}.3^{10}-2^{10}.3^9}{2^9.3^{10}}=\frac{2^{10}.3^9\left(3-1\right)}{2^9.3^{10}}=2\)
d(\(\frac{5^{11}.7^{12}+5^{11}.7^{11}}{5^{12}.7^{12}+9.5^{11}.7^{11}}=\frac{5^{11}.7^{11}\left(7+1\right)}{5^{11}.7^{11}\left(35+9\right)}=\frac{1}{6}\)
\(\dfrac{-5}{12}\times\dfrac{2}{11}+\dfrac{-5}{11}\times\dfrac{9}{11}+\dfrac{5}{12}\\ =\dfrac{5}{12}\times\left(1-\dfrac{2}{11}\right)+\dfrac{-5}{11}\times\dfrac{9}{11}\\ =\dfrac{5}{12}\times\dfrac{9}{11}+\dfrac{-5}{11}\times\dfrac{9}{11}\\ =\dfrac{9}{11}\times\left(\dfrac{5}{12}-\dfrac{5}{11}\right)\\ =\dfrac{9}{11}\times\dfrac{-5}{132}=-\dfrac{15}{484}\)