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Bai 1: tính nhanh A) -5/9 + 3/5 - 3/9 + -2/5 B) -5/13 + (3/5 + 3/1 - 4/10) C) 5/17 - 9/15 - 2/-17 + -2/15 D) (1/9 - 9/17) + 3/6 - ( 12/17 - 1/2) + -1/9 Bài 5: tính tổng A) 1/3 + -1/4 + 1/5 + 1/-6 + -1/-7 + 1/6 + -1/5 + 1/4 + 1/3 B) 1/12 +1/2.3+1/3.4+..+1/99100 Giúp mình nhé nhanh
c: Ta có: \(-\dfrac{5}{13}-\left(\dfrac{3}{5}+\dfrac{3}{13}-\dfrac{4}{10}\right)\)
\(=\dfrac{-5}{13}-\dfrac{3}{5}-\dfrac{3}{13}+\dfrac{2}{5}\)
\(=\dfrac{-8}{13}-\dfrac{1}{5}\)
\(=\dfrac{-53}{65}\)
d: Ta có: \(\left(\dfrac{1}{9}-\dfrac{9}{17}\right)+\dfrac{3}{6}-\left(\dfrac{12}{17}-\dfrac{1}{2}\right)+\dfrac{5}{9}\)
\(=\dfrac{1}{9}-\dfrac{9}{17}+\dfrac{1}{2}-\dfrac{12}{17}+\dfrac{1}{2}+\dfrac{5}{9}\)
\(=\dfrac{2}{3}+1-\dfrac{21}{17}\)
\(=\dfrac{22}{51}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Bạn tính hai vế à.!? Hay tính vế thứ nhất rồi với vế thứ 2.!???
![](https://rs.olm.vn/images/avt/0.png?1311)
o: \(\dfrac{\left(-1\right)^6\cdot3^5\cdot4^3}{9^2\cdot2^5}=\dfrac{3^5\cdot2^6}{2^5\cdot3^4}=\dfrac{3^5}{3^4}\cdot\dfrac{2^6}{2^5}=3\cdot2=6\)
s: \(\dfrac{\dfrac{2}{7}+\dfrac{2}{5}+\dfrac{2}{17}-\dfrac{2}{25}}{\dfrac{3}{14}+\dfrac{3}{10}+\dfrac{3}{34}-\dfrac{3}{50}}\)
\(=\dfrac{2\left(\dfrac{1}{7}+\dfrac{1}{5}+\dfrac{1}{17}-\dfrac{1}{25}\right)}{\dfrac{3}{2}\left(\dfrac{1}{7}+\dfrac{1}{5}+\dfrac{1}{17}-\dfrac{1}{25}\right)}\)
\(=2:\dfrac{3}{2}=\dfrac{4}{3}\)
t: \(\sqrt{\dfrac{4}{9}}-\dfrac{1}{2}:\left|-\dfrac{2}{3}\right|\)
\(=\dfrac{2}{3}-\dfrac{1}{2}:\dfrac{2}{3}\)
\(=\dfrac{2}{3}-\dfrac{3}{4}=\dfrac{8-9}{12}=-\dfrac{1}{12}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
a: \(A=\dfrac{3^6\cdot3^8\cdot5^4-3^{13}\cdot5^{13}\cdot5^{-9}}{3^{12}\cdot5^6+5^6\cdot3^{12}}\)
\(=\dfrac{3^{14}\cdot5^4-3^{13}\cdot5^4}{2\cdot3^{12}\cdot5^6}\)
\(=\dfrac{3^{13}\cdot5^4\cdot\left(3-1\right)}{2\cdot3^{12}\cdot5^6}=\dfrac{3}{5^2}=\dfrac{3}{25}\)
c: \(C=\dfrac{\dfrac{27}{64}+\dfrac{125}{64}-5\cdot\dfrac{16-15}{12}}{\dfrac{25}{64}+\dfrac{4}{9}-\dfrac{5}{6}}\)
\(=\dfrac{47}{24}:\dfrac{1}{576}=47\cdot24=1128\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(\dfrac{1}{19}+\dfrac{9}{19\cdot29}+...+\dfrac{9}{1999\cdot2009}\)
\(=\dfrac{1}{19}+\dfrac{9}{10}\left(\dfrac{10}{19\cdot29}+...+\dfrac{10}{1999\cdot2009}\right)\)
\(=\dfrac{1}{19}+\dfrac{9}{10}\left(\dfrac{1}{19}-\dfrac{1}{2009}\right)\)
\(=\dfrac{1}{19}+\dfrac{1791}{38171}=\dfrac{200}{2009}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(\dfrac{17}{-26}\cdot\left(\dfrac{1}{6}-\dfrac{5}{3}\right):\dfrac{17}{13}-\dfrac{20}{3}\cdot\left(\dfrac{2}{5}-\dfrac{1}{4}\right)+\dfrac{2}{3}\left(\dfrac{6}{5}-\dfrac{9}{2}\right)\)
\(=\dfrac{-17}{26}\cdot\dfrac{13}{17}\cdot\dfrac{1-10}{6}-\dfrac{20}{3}\cdot\dfrac{8-5}{20}+\dfrac{2}{3}\cdot\dfrac{12-45}{10}\)
\(=\dfrac{-1}{2}\cdot\dfrac{-9}{6}-\dfrac{20}{3}\cdot\dfrac{3}{20}+\dfrac{2}{3}\cdot\dfrac{-33}{10}\)
\(=\dfrac{3}{4}-1-\dfrac{66}{30}=\dfrac{-1}{4}-\dfrac{11}{5}=\dfrac{-5-44}{20}=-\dfrac{49}{20}\)
thầy @phynit sửa chỗ \(\left(BC-AB^2\right)\) thành \(\left(BC-AB\right)^2\) giúp em với ạ!
bài 1, 2b, 3a, 5b em lm đúng mà, s đc 6 nhể, trình bày sai chỗ nìu ạ