\(\left[\dfrac{3}{7}.\dfrac{4}{15}+\dfrac{1}{3}.\left(9^{15}\right)\right]^0.\dfrac{1}{3}.\dfrac{68}{124}\)
Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
[3/7.4/15+1/3.(9^15)]^0.1/3.6^8/12^4
= 1.1/3.(2.3)^8/(3.4)^4
= 1/3.2^8.3^8/3^4.4^4
= 1/3.2^8.3^8/3^4.2^8
= 1/3.3^8/3^4
= 1/3.3^4=27
(dấu . là nhân nha)
1: \(=\dfrac{16}{15}\left(-\dfrac{4}{9}+\dfrac{3}{7}\right)+\dfrac{16}{15}\left(\dfrac{4}{7}-\dfrac{5}{9}\right)\)
\(=\dfrac{16}{15}\left(-\dfrac{4}{9}+\dfrac{3}{7}+\dfrac{4}{7}-\dfrac{5}{9}\right)=0\)
2: \(=\dfrac{29}{9}\left(15+\dfrac{4}{7}-8-\dfrac{1}{7}+\dfrac{15}{7}-\dfrac{1}{7}\right)\)
\(=\dfrac{20}{9}\cdot\left(7\cdot\dfrac{18}{7}\right)=\dfrac{20}{9}\cdot18=40\)
a: =-8/28-7/28=-15/28
b: \(=\dfrac{-4}{18}+\dfrac{3}{7}\cdot\dfrac{14}{15}=\dfrac{-2}{9}+\dfrac{14}{15}=\dfrac{-10+42}{45}=\dfrac{32}{45}\)
c: \(=\dfrac{-3\cdot5+7\cdot2}{20}\cdot\dfrac{-5}{1}-\dfrac{2}{9}\)
\(=\dfrac{-7}{4}-\dfrac{2}{9}=\dfrac{-63}{36}-\dfrac{8}{36}=-\dfrac{71}{36}\)
a/ \(\dfrac{\left(1+2+.....+100\right)\left(\dfrac{1}{3}-\dfrac{1}{5}-\dfrac{1}{7}-\dfrac{1}{9}\right)\left(6,3.12-21.36\right)}{\dfrac{1}{2}+\dfrac{1}{3}+.......+\dfrac{1}{100}}\)
\(=\dfrac{\left(1+2+3+.....+100\right)\left(\dfrac{1}{3}-\dfrac{1}{5}-\dfrac{1}{7}-\dfrac{1}{9}\right).0}{\dfrac{1}{2}+\dfrac{1}{3}+.......+\dfrac{1}{100}}\)
\(=\dfrac{0}{\dfrac{1}{2}+\dfrac{1}{3}+.....+\dfrac{1}{100}}\)
\(=0\)
a) Ta có: \(\left(\dfrac{9}{25}-2\cdot18\right):\left(3\dfrac{4}{5}+0.2\right)\)
\(=\left(\dfrac{9}{25}-36\right):\left(\dfrac{19}{5}+\dfrac{1}{5}\right)\)
\(=\left(\dfrac{9}{25}-\dfrac{900}{25}\right):\dfrac{20}{5}\)
\(=\dfrac{-891}{25}\cdot\dfrac{1}{4}\)
\(=-\dfrac{891}{100}\)
b) Ta có: \(\dfrac{3}{8}\cdot19\dfrac{1}{3}+\dfrac{3}{8}\cdot33\dfrac{1}{3}\)
\(=\dfrac{3}{8}\cdot\dfrac{58}{3}+\dfrac{3}{8}\cdot\dfrac{100}{3}\)
\(=\dfrac{58}{8}+\dfrac{100}{8}\)
\(=\dfrac{158}{8}=\dfrac{79}{4}\)
c) Ta có: \(15\cdot\left(-\dfrac{2}{3}\right)^2-\dfrac{7}{3}\)
\(=15\cdot\dfrac{4}{9}-\dfrac{7}{3}\)
\(=\dfrac{20}{3}-\dfrac{7}{3}\)
\(=\dfrac{13}{3}\)
d) Ta có: \(\dfrac{1}{2}\sqrt{64}-\sqrt{\dfrac{4}{25}}+\left(-1\right)^{2007}\)
\(=\dfrac{1}{2}\cdot8-\dfrac{2}{5}-1\)
\(=4-1-\dfrac{2}{5}\)
\(=3-\dfrac{2}{5}\)
\(=\dfrac{15}{5}-\dfrac{2}{5}=\dfrac{13}{5}\)
e) Ta có: \(\left(-\dfrac{5}{2}\right)^2:\left(-15\right)-\left(0.45+\dfrac{3}{4}\right)\cdot\left(-1\dfrac{5}{9}\right)\)
\(=\dfrac{25}{4}\cdot\dfrac{-1}{15}-\left(\dfrac{9}{20}+\dfrac{15}{20}\right)\cdot\dfrac{-14}{9}\)
\(=\dfrac{-25}{60}-\dfrac{24}{20}\cdot\dfrac{-14}{9}\)
\(=\dfrac{-25}{60}+\dfrac{28}{15}\)
\(=\dfrac{-25}{60}+\dfrac{112}{60}\)
\(=\dfrac{87}{60}=\dfrac{29}{20}\)
f) Ta có: \(\left(-\dfrac{1}{3}\right)-\left(-\dfrac{3}{5}\right)^0+\left(1-\dfrac{1}{2}\right)^2:2\)
\(=-\dfrac{1}{3}-1+\left(\dfrac{1}{2}\right)^2\cdot\dfrac{1}{2}\)
\(=\dfrac{-4}{3}+\dfrac{1}{4}\cdot\dfrac{1}{2}\)
\(=\dfrac{-4}{3}+\dfrac{1}{8}\)
\(=\dfrac{-32}{24}+\dfrac{3}{24}=\dfrac{-29}{24}\)
g) Ta có: \(\left(\dfrac{1}{2}\right)^{15}\cdot\left(\dfrac{1}{4}\right)^{20}\)
\(=\left(\dfrac{1}{2}\right)^{15}\cdot\left(\dfrac{1}{2}\right)^{40}\)
\(=\left(\dfrac{1}{2}\right)^{55}\)
\(=\dfrac{1}{2^{55}}\)
h) Ta có: \(\dfrac{5^4\cdot20}{25^5\cdot4^5}\)
\(=\dfrac{5^4\cdot5\cdot2^2}{5^{10}\cdot2^{10}}\)
\(=\dfrac{5^5}{5^{10}}\cdot\dfrac{2^2}{2^{10}}\)
\(=\dfrac{1}{5^5}\cdot\dfrac{1}{2^8}\)
\(=\dfrac{1}{800000}\)
2: \(=\dfrac{203}{60}\cdot\dfrac{81}{1225}=\dfrac{783}{3500}\)
A -\(\dfrac{24}{25}\)
B -\(\dfrac{5}{21}\)
C -\(\dfrac{24}{47}\)
D -\(\dfrac{19}{42}\)
tick cho mk
a, \(\left(\dfrac{-2}{3}+\dfrac{3}{7}\right)-\dfrac{5}{21}:\dfrac{4}{5}+\left(\dfrac{-1}{3}+\dfrac{4}{7}\right):\dfrac{4}{5}\\ = -\dfrac{5}{21}:\dfrac{4}{5}+ \left(-\dfrac{5}{21}\right):\dfrac{4}{5}\\ =\left[-\dfrac{5}{21}+\left(-\dfrac{5}{21}\right)\right]:\dfrac{4}{5}\\ -\dfrac{10}{21}:\dfrac{4}{5}\\ =-\dfrac{25}{42}\)
b,
\(\dfrac{5}{9}:\left(\dfrac{1}{11}-\dfrac{5}{22}\right)+\dfrac{5}{9}:\left(\dfrac{1}{15}-\dfrac{2}{3}\right)\\ =\dfrac{5}{9}:\dfrac{-3}{22}+\dfrac{5}{9}:-\dfrac{3}{5}\\ =\dfrac{5}{9}:\left(\dfrac{-3}{22}+-\dfrac{3}{5}\right)\\ =\dfrac{5}{9}:-\dfrac{81}{110}\\ =-\dfrac{550}{729}\)
\(\left[\dfrac{3}{7}\times\dfrac{4}{15}+\dfrac{1}{3}\times\left(9^{15}\right)\right]^0\times\dfrac{1}{3}\times\dfrac{68}{124}\)
\(=1\times\dfrac{1}{3}\times\dfrac{17}{31}\)
\(=\dfrac{1}{3}\times\dfrac{17}{31}\)
\(=\dfrac{17}{93}\)
\(\left[\dfrac{3}{7}.\dfrac{4}{15}+\dfrac{1}{3}.\left(9^{15}\right)\right].\dfrac{1}{3}.\dfrac{68}{124}\)
= \(1.\dfrac{1}{3}.\dfrac{68}{124}\)
= \(\dfrac{1}{3}.\dfrac{68}{124}=\dfrac{17}{93}\)