\(\left(\frac{2}{3}\right)^3\):\(\left(\frac{-1}{3}...">
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7 tháng 3 2018

\(C=\frac{8}{9}.\frac{15}{16}.\frac{24}{25}.........\frac{2499}{2500}\)

\(=\frac{2.4}{3^2}.\frac{3.5}{4^2}.\frac{4.6}{5^2}......\frac{49.51}{50^2}\)

\(=\frac{2.3.4....49}{3.4.5....50}.\frac{4.5.6....51}{3.4.5....50}\)

\(=\frac{1}{25}.17=\frac{17}{25}\)

7 tháng 3 2018

\(a)\) \(A=\left(1-\frac{1}{2}\right)\left(1-\frac{1}{3}\right)\left(1-\frac{1}{4}\right).....\left(1-\frac{1}{1000}\right)\)

\(A=\frac{1}{2}.\frac{2}{3}.\frac{3}{4}.....\frac{999}{1000}\)

\(A=\frac{1.2.3.....999}{2.3.4.....1000}\)

\(A=\frac{1}{1000}.\frac{2.3.4.....999}{2.3.4.....999}\)

\(A=\frac{1}{1000}\)

Vậy \(A=\frac{1}{1000}\)

9 tháng 7 2021

1. 

a.\(\left(\frac{1}{2}\right)^2=\frac{1}{4}\)

b. \(\left(\frac{1}{2}\right)^3=\frac{1}{8}\)

c. \(\left(\frac{-3}{5}\right)^5=\frac{-243}{3125}\)

d. \(\left(\frac{-1}{5}\right)^2=\frac{1}{25}\)

e. \(\left(\frac{-1}{6}\right)^3=\frac{-1}{216}\)

10 tháng 7 2021

Trả lời:

Bài 1: 

a, \(\left(\frac{1}{2}\right)^4=\frac{1^4}{2^4}=\frac{1}{16}\)

b, \(\left(\frac{1}{2}\right)^3=\frac{1^3}{2^3}=\frac{1}{8}\)

c, \(\left(\frac{-3}{5}\right)^2=\frac{\left(-3\right)^2}{5^2}=\frac{9}{25}\)

d, \(\left(\frac{-1}{5}\right)^2=\frac{\left(-1\right)^2}{5^2}=\frac{1}{25}\)

e, \(\left(\frac{-1}{6}\right)^3=\frac{\left(-1\right)^3}{6^3}=\frac{-1}{216}\)

Bài 2:

a, \(\left(\frac{3}{2}\right)^2.\left(\frac{4}{3}\right)^2=\frac{9}{4}.\frac{16}{9}=4\)

b, \(\left(-\frac{1}{2}\right)^3.\left(\frac{2}{3}\right)^3=-\frac{1}{8}.\frac{8}{27}=-\frac{1}{27}\)

c, \(\left(-\frac{1}{2}\right)^2.\left(\frac{2}{5}\right)^2=\frac{1}{4}.\frac{4}{25}=\frac{1}{25}\)

d, \(\left(-\frac{1}{2}\right)^3.\left(\frac{2}{3}\right)^3=-\frac{1}{8}.\frac{8}{27}=-\frac{1}{27}\)

e, \(\left(-5\right)^3.\frac{1}{5}=-125.\frac{1}{5}=-25\)

f, \(\left(\frac{2}{9}\right)^5.\left(-\frac{27}{4}\right)^5=\frac{2^5}{9^5}.\frac{\left(-27\right)^5}{4^5}=\frac{2^5.\left(-27\right)^5}{9^5.4^5}=\frac{2^5.\left[\left(-3\right)^3\right]^5}{\left(3^2\right)^5.\left(2^2\right)^5}=-\frac{2^5.3^{15}}{3^{10}.2^{10}}=\frac{3^5}{2^5}\)

a) \(\left(-\frac{1}{4}\right)^0=1\)

b) \(\left(-2\frac{1}{3}\right)^2=\left(-\frac{7}{3}\right)^2=\frac{49}{9}\)

c) \(\left(\frac{4}{5}\right)^{-2}=\frac{25}{16}\)

d) \(\left(0,5\right)^{-3}=8\)

e) \(\left(-1\frac{1}{3}\right)^4=\left(-\frac{4}{3}\right)^4=\frac{256}{81}\)

8 tháng 8 2019

a, \(\left(\frac{-1}{4}\right)^0\) = 1

Bất kỳ số nguyên nào nếu có mũ bằng 0 đều bằng 1

b, \(\left(-2\frac{1}{3}\right)^2=\left(-\frac{7}{3}\right)^2=\frac{49}{9}\)

a) Ta có: \(\frac{-1}{12}-\left(2\frac{5}{8}-\frac{1}{3}\right)\)

\(=-\frac{1}{12}-\frac{21}{8}+\frac{1}{3}\)

\(=\frac{-6}{72}-\frac{189}{72}+\frac{24}{72}\)

\(=-\frac{19}{8}\)

b) Ta có: \(-1,75-\left(\frac{-1}{9}-2\frac{1}{18}\right)\)

\(=\frac{-7}{4}+\frac{1}{9}+\frac{37}{18}\)

\(=\frac{-63}{36}+\frac{4}{36}+\frac{74}{36}\)

\(=\frac{5}{12}\)

c) Ta có: \(\frac{2}{5}+\frac{-4}{3}+\frac{-1}{2}\)

\(=\frac{12}{30}+\frac{-40}{30}+\frac{-15}{30}\)

\(=-\frac{43}{30}\)

d) Ta có: \(\frac{3}{12}-\left(\frac{6}{15}-\frac{3}{10}\right)\)

\(=\frac{3}{12}-\frac{6}{15}+\frac{3}{10}\)

\(=\frac{15}{60}-\frac{24}{60}+\frac{18}{60}\)

\(=\frac{3}{20}\)

e) Ta có: \(\left(8\frac{5}{11}+3\frac{5}{8}\right)-3\frac{5}{11}\)

\(=\frac{93}{11}+\frac{29}{8}-\frac{38}{11}\)

\(=5+\frac{29}{8}=\frac{40}{8}+\frac{29}{8}=\frac{69}{8}\)

f) Ta có: \(\frac{4}{9}:\left(-\frac{1}{7}\right)+6\frac{5}{9}:\left(-\frac{1}{7}\right)\)

\(=\frac{4}{9}\cdot\left(-7\right)+\frac{59}{9}\cdot\left(-7\right)\)

\(=\left(-7\right)\cdot\left(\frac{4}{9}+\frac{59}{9}\right)=\left(-7\right)\cdot7=-49\)

g) Ta có: \(\frac{-1}{4}\cdot13\frac{9}{11}-0,25\cdot6\frac{2}{11}\)

\(=\frac{-1}{4}\cdot\frac{152}{11}+\frac{-1}{4}\cdot\frac{68}{11}\)

\(=\frac{-1}{4}\cdot\left(\frac{152}{11}+\frac{68}{11}\right)=-\frac{1}{4}\cdot20=-5\)

h) Ta có: \(5\frac{27}{5}+\frac{27}{23}+0,5-\frac{5}{27}+\frac{16}{23}\)

\(=\frac{52}{5}+\frac{27}{23}+\frac{1}{2}-\frac{5}{27}+\frac{16}{23}\)

\(=\frac{52}{5}+\frac{43}{23}+\frac{1}{2}-\frac{5}{27}\)

\(=\frac{64584}{6210}+\frac{11610}{6210}+\frac{3105}{6210}-\frac{1150}{6210}\)

\(=\frac{78149}{6210}\)

i) Ta có: \(\frac{3}{8}\cdot27\frac{1}{5}-51\frac{1}{5}\cdot\frac{3}{8}+19\)

\(=\frac{3}{8}\cdot\frac{136}{5}-\frac{3}{8}\cdot\frac{206}{5}+\frac{3}{8}\cdot\frac{152}{3}\)

\(=\frac{3}{8}\cdot\left(\frac{136}{5}-\frac{206}{5}+\frac{152}{3}\right)=\frac{3}{8}\cdot\frac{110}{3}\)

\(=\frac{55}{4}\)

a) \(\frac{-6}{21}.\frac{3}{2}=-\frac{3}{7}\)          b) \(\left(-3\right).\left(\frac{-7}{12}\right)=\frac{21}{12}=\frac{7}{4}\)

c) \(\left(\frac{11}{12}:\frac{33}{16}\right).\frac{3}{5}=\frac{11}{12}.\frac{16}{33}.\frac{3}{5}=\frac{4}{15}\)

d) \(\sqrt{\left(-7\right)^2}+\sqrt{\frac{2}{16}}=7+\sqrt{\frac{1}{8}}\)

c) \(\frac{1}{2}.\sqrt{100}-\sqrt{\frac{1}{16}}+\left(\frac{1}{3}\right)^0=\frac{1}{2}.10-\frac{1}{4}+1=5\frac{3}{4}\)

1 tháng 8 2020

spam. cho mình xóa nhaa :3