Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
![](https://rs.olm.vn/images/avt/0.png?1311)
a) Ta có: \(\left(\dfrac{617}{191}+\dfrac{29}{33}-\dfrac{115}{17}\right)\cdot\left(\dfrac{1}{4}-\dfrac{1}{5}-\dfrac{1}{20}\right)\)
\(=\left(\dfrac{617}{191}+\dfrac{29}{33}-\dfrac{115}{17}\right)\cdot\left(\dfrac{5}{20}-\dfrac{4}{20}-\dfrac{1}{20}\right)\)
\(=0\cdot\left(\dfrac{617}{191}+\dfrac{29}{33}-\dfrac{115}{17}\right)=0\)
b) Ta có: \(\dfrac{12}{5}\cdot\left(\dfrac{10}{3}-\dfrac{5}{12}\right)\)
\(=\dfrac{12}{5}\cdot\left(\dfrac{40}{12}-\dfrac{5}{12}\right)\)
\(=\dfrac{12}{5}\cdot\dfrac{35}{12}\)
=7
![](https://rs.olm.vn/images/avt/0.png?1311)
(617/191+29/33-115/117)*(1/4-1/5-1/20)
= (617/191+29/33-115/117)*(5/20-4/20-1/20)
=(617/191+29/33-115/117)*0
=0
![](https://rs.olm.vn/images/avt/0.png?1311)
a) Ta có: \(\dfrac{-5}{18}+\dfrac{32}{45}-\dfrac{9}{10}\)
\(=\dfrac{-25}{90}+\dfrac{64}{90}-\dfrac{81}{90}\)
\(=\dfrac{-42}{90}=-\dfrac{7}{15}\)
b) Ta có: \(\left(-\dfrac{1}{4}+\dfrac{51}{33}-\dfrac{5}{3}\right)-\left(-\dfrac{15}{12}+\dfrac{6}{11}-\dfrac{42}{29}\right)\)
\(=\dfrac{-1}{4}+\dfrac{17}{11}-\dfrac{5}{3}+\dfrac{5}{4}-\dfrac{6}{11}+\dfrac{42}{29}\)
\(=\dfrac{-5}{3}+\dfrac{42}{29}\)
\(=\dfrac{-145}{87}+\dfrac{126}{87}=\dfrac{-19}{87}\)
c) Ta có: \(1-\dfrac{1}{2}+2-\dfrac{2}{3}+3-\dfrac{3}{4}+4-\dfrac{1}{4}-3-\dfrac{1}{3}-2-\dfrac{1}{2}-1\)
\(=\left(1-1\right)-\left(\dfrac{1}{2}+\dfrac{1}{2}\right)+\left(2-2\right)-\left(\dfrac{2}{3}+\dfrac{1}{3}\right)+\left(3-3\right)-\left(\dfrac{3}{4}+\dfrac{1}{4}\right)+4\)
\(=-1-1-1+4\)
=1
a) Ta có: −518+3245−910−518+3245−910
=−2590+6490−8190=−2590+6490−8190
=−4290=−715=−4290=−715
b) Ta có: (−14+5133−53)−(−1512+611−4229)(−14+5133−53)−(−1512+611−4229)
=−14+1711−53+54−611+4229=−14+1711−53+54−611+4229
=−53+4229=−53+4229
=−14587+12687=−1987=−14587+12687=−1987
c) Ta có: 1−12+2−23+3−34+4−14−3−13−2−12−11−12+2−23+3−34+4−14−3−13−2−12−1
=(1−1)−(12+12)+(2−2)−(23+13)+(3−3)−(34+14)+4=(1−1)−(12+12)+(2−2)−(23+13)+(3−3)−(34+14)+4
=−1−1−1+4=−1−1−1+4
=1
![](https://rs.olm.vn/images/avt/0.png?1311)
\(\left(\dfrac{67}{111}+\dfrac{2}{33}-\dfrac{15}{117}\right).\left(\dfrac{1}{3}-\dfrac{1}{4}-\dfrac{1}{12}\right)\)
\(=\left(\dfrac{67}{111}+\dfrac{2}{33}-\dfrac{15}{117}\right).\left(\dfrac{4}{12}-\dfrac{3}{12}-\dfrac{1}{12}\right)\)
\(=\left(\dfrac{67}{111}+\dfrac{2}{33}-\dfrac{15}{117}\right).\left(\dfrac{4-3-1}{12}\right)\)
\(=\left(\dfrac{66}{111}+\dfrac{2}{33}+\dfrac{15}{117}\right).0\)
\(=0\)
\(\left(\dfrac{67}{111}+\dfrac{2}{33}-\dfrac{15}{117}\right)\cdot\left(\dfrac{1}{3}-\dfrac{1}{4}-\dfrac{1}{12}\right)\\ =\left(\dfrac{67}{111}+\dfrac{2}{33}-\dfrac{15}{117}\right)\cdot\left(\dfrac{4}{12}-\dfrac{3}{12}-\dfrac{1}{12}\right)\\= \left(\dfrac{67}{111}+\dfrac{2}{33}-\dfrac{15}{117}\right)\cdot\left(\dfrac{4-3-1}{12}\right)\\= \left(\dfrac{67}{111}+\dfrac{2}{33}-\dfrac{15}{117}\right)\cdot0\\ =0\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Gợi ý: Sử dụng tính chất phân phối của phép nhân đối với phép cộng để nhóm thừa số chung ra ngoài.
![](https://rs.olm.vn/images/avt/0.png?1311)
\(1,A=-\dfrac{3}{4}.\left(0,125-1\dfrac{1}{2}\right):\dfrac{33}{16}-25\%\)
\(A=-\dfrac{3}{4}.\left(0,125-\dfrac{3}{2}\right):\dfrac{33}{16}-\dfrac{1}{4}\)
\(A=-\dfrac{3}{4}.\left(-\dfrac{11}{8}\right):\dfrac{33}{16}-\dfrac{1}{4}\)
\(A=\dfrac{33}{32}:\dfrac{33}{16}-\dfrac{1}{4}\)
\(A=\dfrac{33}{32}.\dfrac{16}{33}-\dfrac{1}{4}\)
\(A=\dfrac{1}{2}-\dfrac{1}{4}\)
\(A=\dfrac{2}{4}-\dfrac{1}{4}\)
\(A=\dfrac{1}{4}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
a) \(\dfrac{-1}{20}=\dfrac{-7}{140}\)
\(\dfrac{5}{7}=\dfrac{100}{140}\)
mà -7<100
nên \(-\dfrac{1}{20}< \dfrac{5}{7}\)
b) \(\dfrac{216}{217}< 1\)
\(1< \dfrac{1164}{1163}\)
nên \(\dfrac{216}{217}< \dfrac{1164}{1163}\)
c) \(\dfrac{-12}{17}=\dfrac{-180}{255}\)
\(\dfrac{-14}{15}=\dfrac{-238}{255}\)
mà -180>-238
nên \(-\dfrac{12}{17}>\dfrac{-14}{15}\)
d) \(\dfrac{27}{29}>0\)
\(0>-\dfrac{2727}{2929}\)
nên \(\dfrac{27}{29}>-\dfrac{2727}{2929}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(M=\dfrac{120-\dfrac{1}{2}.40.5.\dfrac{1}{5}.20.\dfrac{1}{4}-20}{1+5+....+41}\\ =\dfrac{120-20.5-20}{1+5+...+41}\\ =\dfrac{0}{1+5+...+41}\\ =0\)
\(N=10101\left(\dfrac{6}{3.7.11.13.37}+\dfrac{6}{2.3.7.11.13.37}-\dfrac{7}{3.7.11.13.37}\right)\\ =10101\left(-\dfrac{2}{2.3.7.11.13.37}+\dfrac{6}{2.3.7.11.13.37}\right)\\ =3.7.13.37\left(\dfrac{4}{2.3.7.11.13.37}\right)=\dfrac{4}{2.11}=\dfrac{2}{11}\)