(mình đag cần gấp)
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theo cách cấp 2
A=(10001x2012+1)x(10001x2013-1) - 100012x2012x2013
=100012x 2012 x 2013 - 2012 x 10001 + 10001 x 2013 - 1 - 100012 x 2012 x 2013
= 10001 x ( 2012 + 2013 )-1
=40254025 - 1
= 40254024
Đúng hay không tui không biết ha
![](https://rs.olm.vn/images/avt/0.png?1311)
a: =-5/21
b: \(=\dfrac{15}{4}-\dfrac{9}{4}=\dfrac{6}{4}=\dfrac{3}{2}\)
c: \(=\left(\dfrac{31}{10}-\dfrac{25}{10}\right)\cdot3-2=\dfrac{6}{10}\cdot3-2=\dfrac{18}{10}-\dfrac{20}{10}=\dfrac{-2}{10}=\dfrac{-1}{5}\)
d: \(=\dfrac{4}{3}\cdot\dfrac{1}{20}=\dfrac{4}{60}=\dfrac{1}{15}\)
a: \(=\dfrac{-15}{25}\cdot\dfrac{8}{16}=\dfrac{-3}{5}\cdot\dfrac{1}{2}=\dfrac{-3}{10}\)
b: \(=\dfrac{2}{3}+\dfrac{2}{7}=\dfrac{20}{21}\)
c: \(=\dfrac{3}{2}\cdot\dfrac{-3+2}{4}=\dfrac{3}{2}\cdot\dfrac{-1}{4}=\dfrac{-3}{8}\)
d: \(=\dfrac{7}{11}\cdot\dfrac{11}{7}\cdot\dfrac{-3}{41}=-\dfrac{3}{41}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(a,\dfrac{10}{11}+\dfrac{3}{11}:3-\dfrac{1}{7}=\dfrac{10}{11}+\dfrac{3}{11}.\dfrac{1}{3}-\dfrac{1}{7}=\dfrac{10}{11}+\dfrac{1}{11}-\dfrac{1}{7}=1-\dfrac{1}{7}=\dfrac{6}{7}\\ b,\dfrac{-3}{7}+\dfrac{5}{13}+\dfrac{3}{7}=\left(\dfrac{-3}{7}+\dfrac{3}{7}\right)+\dfrac{5}{13}=\dfrac{5}{13}\\ c,-\dfrac{3}{13}+\dfrac{7}{12}-\dfrac{10}{13}+\dfrac{5}{12}=\left(-\dfrac{3}{13}-\dfrac{10}{13}\right)+\left(\dfrac{7}{12}+\dfrac{5}{12}\right)=\left(-1\right)+1=0\)
\(d,\dfrac{5}{3}.\dfrac{7}{25}+\dfrac{5}{3}.\dfrac{21}{25}-\dfrac{5}{3}.\dfrac{7}{25}=\left(\dfrac{5}{3}.\dfrac{7}{25}-\dfrac{5}{3}.\dfrac{7}{25}\right)+\dfrac{5}{3}.\dfrac{3.7}{5.5}=\dfrac{7}{5}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
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1c
2c
3b
4d
5b
6d
7a