\(a.\frac{4^3.1^5}{9^2.8^2}=\frac{2^6.1}{3^4.2^6}=\frac{1}{81}\)
\(b.\frac{25^2.2^2.4^1}{3^3.2^3.224}=\frac{5^4.2^4}{3^3.2^8.7}=\frac{5^4}{3^3.2^4.7}=\frac{625}{3024}\)
\(c.\frac{25^3.5^6.5^2}{9^4.10^2}=\frac{5^{14}}{3^8.2^2.5^2}=\frac{5^{10}}{3^8.2^2}\)
tính:a.\(2^6.5^6\)b.\(8^2.5^2\)c.\(4^3.5^3\)d.\(5^2.6^2.3^2\)e.\(\frac{625^5}{25^8}\)g.\(\frac{3^9}{7}.\frac{7^9}{3}\)
giúp mình nha các bạn !
Thực hiện phép tính:
a) \(\frac{-5}{9}.\left(\frac{3}{10}-\frac{2}{5}\right)\)
b) \(2^8:2^5+3^3.2-12\)
c) \(\frac{1}{2}\sqrt{64}-\sqrt{\frac{4}{25}}+1^{2012}\)
d) \(\left(-3\right)^2+\sqrt{\frac{16}{25}}-\sqrt{9}+\frac{\sqrt{81}}{3}\)
Chứng minh rằng:
\(\frac{3}{1^2.2^2}+\frac{5}{2^2.4^2}+\frac{7}{3^2.4^2}+...+\frac{19}{9^2.10^2}<1\)
chứng minh rằng:
a) A= \(\frac{3}{1^2.2^2}+\frac{5}{2^2.3^2}+\frac{7}{3^2.4^2}+...+\frac{19}{9^2.10^2}\)<1
b)B=\(\frac{1}{3}+\frac{2}{3^2}+\frac{3}{3^3}+\frac{4}{4^4}+...+\frac{100}{3^{100}}< \frac{3}{4}\)
CMR: \(\frac{3}{1^2.2^2}+\frac{5}{2^2.3^2}+\frac{7}{3^2.4^2}+...+\frac{19}{9^2.10^2}< 1\)
CMR : \(\frac{3}{1^2.2^2}+\frac{5}{2^2.3^2}+\frac{7}{3^2.4^2}+...+\frac{19}{9^2.10^2}< 1\)
Bài 1:Tính
a) \((\frac{1}{2})^3.\left(\frac{1}{4}\right)^2\)
b)\(27^3:9^3\)
c)\(125^2:25^3\)
d)\(\frac{27^2.8^5}{6^6.32^3}\)
Chướng minh rằng:
a, \(\frac{1}{1^2.2^2}\)+$\frac{5}{2^2.3^2}$+$\frac{5}{3^2.4^2}$+...+$\frac{5}{9^2.10^2}$ <1
b, \(\frac{1}{3}\)+\(\frac{2}{3^2}\)+$\frac{3}{3^3}$+$\frac{4}{3^4}$+...+$\frac{100}{3^100}$ <\(\frac{3}{4}\)
\(\frac{3}{1^2.2^2}+\frac{5}{2^2.3^2}+\frac{7}{3^2.4^2}+..........+\frac{19}{9^2.10^2}<1\)
