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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) \(\left(-1\frac{1}{3}\right)^4=\left(-\frac{4}{3}\right)^4=\frac{256}{81}\)
a, \(\left(\frac{-1}{4}\right)^0\) = 1
Bất kỳ số nguyên nào nếu có mũ bằng 0 đều bằng 1
b, \(\left(-2\frac{1}{3}\right)^2=\left(-\frac{7}{3}\right)^2=\frac{49}{9}\)
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. \(\left(\frac{-1}{6}\right)^3=\frac{-1}{216}\)
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, \(\left(\frac{2}{9}\right)^5.\left(-\frac{27}{4}\right)^5=\frac{2^5}{9^5}.\frac{\left(-27\right)^5}{4^5}=\frac{2^5.\left(-27\right)^5}{9^5.4^5}=\frac{2^5.\left[\left(-3\right)^3\right]^5}{\left(3^2\right)^5.\left(2^2\right)^5}=-\frac{2^5.3^{15}}{3^{10}.2^{10}}=\frac{3^5}{2^5}\)
\(C=\frac{8}{9}.\frac{15}{16}.\frac{24}{25}.........\frac{2499}{2500}\)
\(=\frac{2.4}{3^2}.\frac{3.5}{4^2}.\frac{4.6}{5^2}......\frac{49.51}{50^2}\)
\(=\frac{2.3.4....49}{3.4.5....50}.\frac{4.5.6....51}{3.4.5....50}\)
\(=\frac{1}{25}.17=\frac{17}{25}\)
\(a)\) \(A=\left(1-\frac{1}{2}\right)\left(1-\frac{1}{3}\right)\left(1-\frac{1}{4}\right).....\left(1-\frac{1}{1000}\right)\)
\(A=\frac{1}{2}.\frac{2}{3}.\frac{3}{4}.....\frac{999}{1000}\)
\(A=\frac{1.2.3.....999}{2.3.4.....1000}\)
\(A=\frac{1}{1000}.\frac{2.3.4.....999}{2.3.4.....999}\)
\(A=\frac{1}{1000}\)
Vậy \(A=\frac{1}{1000}\)
\(x = {-b \pm \sqrt{b^2-4ac} \over 2a} đây là biểu thức gì\)
a,\(\frac{31}{1000}\)
b,\(\frac{8}{108}\)
c,0
a,\(\frac{49}{97}\)
b,\(\frac{-1}{4751}\)
a) \(\frac{1}{3}.\frac{-6}{13}.\frac{-9}{10}.\frac{-13}{36}\)
\(=\left(\frac{1}{3}.\frac{-9}{10}\right)\left(\frac{-6}{13}.\frac{-13}{36}\right)\)
\(=\frac{-3}{10}.\frac{1}{6}\)
\(=\frac{-1}{20}\)
b) \(\frac{-1}{3}.\frac{-15}{17}.\frac{34}{45}\)
\(=\frac{-1}{3}.\frac{-2}{3}\)
\(=\frac{2}{9}\)
c) \(\left(1-\frac{1}{5}\right)\left(\frac{-3}{10}+\frac{1}{5}\right)\)
\(=\frac{4}{5}.\frac{-1}{10}\)
\(=\frac{-2}{25}\)
d) \(A=\frac{1}{3}.\frac{4}{5}+\frac{1}{3}.\frac{6}{5}+\frac{2}{3}\)
\(=\frac{1}{3}\left(\frac{4}{5}+\frac{6}{5}\right)+\frac{2}{3}\)
\(=\frac{1}{3}.2+\frac{2}{3}\)
\(=\frac{2}{3}+\frac{2}{3}\)
\(=\frac{4}{3}\)
e) \(11\frac{1}{4}-\left(2\frac{5}{7}+5\frac{1}{4}\right)\)
\(=\left(11\frac{1}{4}-5\frac{1}{4}\right)-2\frac{5}{7}\)
\(=6-2\frac{5}{7}\)
\(=5\frac{7}{7}-2\frac{5}{7}\)
\(=3\frac{2}{7}\)
\(B=\left(1+\frac{1}{2}\right).\left(1+\frac{1}{3}\right).\left(1+\frac{1}{4}\right).....\left(1+\frac{1}{2015}\right)\)
\(B=\frac{3}{2}.\frac{4}{3}.\frac{5}{4}.....\frac{2016}{2015}\)
\(B=\frac{2016}{2}=1008\)
\(C=\frac{3^3.\left(0,5\right)^5}{\left(1,5\right)^4}\)
\(C=\frac{27.\left(\frac{1}{2}\right)^5}{\left(\frac{3}{2}\right)^4}\)
\(C=\frac{27.\frac{1^5}{2^5}}{\frac{3^4}{2^4}}\)
\(C=\frac{27.\frac{1}{32}}{\frac{81}{16}}\)
\(C=\frac{\frac{27}{32}}{\frac{81}{16}}\)
\(C=\frac{27}{32}:\frac{81}{16}\)
\(C=\frac{27}{32}.\frac{16}{81}\)
\(C=\frac{1}{2}.\frac{1}{3}\)
\(C=\frac{1}{6}\)