\(\frac{27^4.4^3}{9^5.8^2}=\frac{\left(3^3\right)^4.\left(2^2\right)^3}{\left(3^2\right)^5.\left(2^3\right)^2}=\frac{3^{12}.2^6}{3^{10}.2^6}=3^2=9\)
=\(\frac{3^{12}.2^6}{3^{15}.2^6}\)
=\(\frac{3^{12}}{3^{12}.3^3}\)
=\(\frac{1}{27}\)
\(\frac{27^4.4^3}{9^5.8^2}=\frac{\left(3^3\right)^4.\left(2^2\right)^3}{\left(3^2\right)^5.\left(2^3\right)^2}=\frac{3^{12}.2^6}{3^{10}.2^6}=3^2=9\)
=\(\frac{3^{12}.2^6}{3^{15}.2^6}\)
=\(\frac{3^{12}}{3^{12}.3^3}\)
=\(\frac{1}{27}\)
\(\dfrac{27^4.4^3}{9^5.8^2}\);\(\dfrac{3^{29}.4^{16}}{27^9.8^{11}}\)
\(A\frac{6^{10}-3^9\cdot2^8\cdot5}{27^3\cdot4^5+16^3\cdot9^4}\)
Thực hiện phép tính
a)\(\frac{27^4\cdot2^3-3^{10}\cdot4^3}{6^4\cdot9^3}\)
b)\(\left(\frac{1}{4\cdot9}+\frac{1}{9\cdot14}+\frac{1}{14\cdot19}+...+\frac{1}{44\cdot49}\right)\)\(\cdot\frac{1-3-5-7-...-49}{89}\)
Tính \(A=\left(\frac{1}{4\cdot9}+\frac{1}{9\cdot14}+\frac{1}{14\cdot19}+...+\frac{1}{44\cdot49}\right)\cdot\frac{1-3-5-7-...-49}{89}\)
\(B=\frac{5\cdot4^{15}\cdot9^9-4\cdot3^{20}\cdot8^9}{5\cdot2^{10}\cdot6^{19}-7\cdot2^{29}\cdot27^6}-\frac{2^{19}\cdot27^3+15\cdot4^9\cdot9^4}{6^9\cdot2^{10}+12^{10}}\)
Chứng minh rằng
a, B = \(\frac{1\cdot2-1}{2!}+\frac{2\cdot3-1}{3!}+\frac{3\cdot4-1}{4!}+....+\frac{99\cdot100-1}{100!}< 2\)
c, C = \(\frac{3}{1^2\cdot2^2}+\frac{5}{2^2\cdot3^2}+\frac{7}{3^2\cdot4^2}+...+\frac{19}{9^2\cdot10^2}< 1\)
The value of \(\frac{9^4\cdot27^3}{81^4}-\frac{2^{11}\cdot4^5}{6^{10}}\) is .........
Rút gọn \(\frac{2^{19}\cdot27^3+15\cdot4^9\cdot9^4}{6^9\cdot2^{10}+12^{10}}\)
Thực hiện các phép tính:
a)A=-4/5:6/5+3/4+1/2-(5/4:1/2)
b)B=2^5.8^4.4^3/16^6+2^3+(1/3)^2+(1/27)^0-3^5.(-25)^2/9^3.5^3
\(\left(\frac{1}{1\cdot2}+\frac{1}{3\cdot4}+...+\frac{1}{49\cdot50}\right)\): \(\left(\frac{2}{26}+\frac{2}{27}+...+\frac{2}{50}\right)\)