1/9.27ⁿn=3ⁿ

1/9=3ⁿ:27ⁿ

1/9=(1/9)ⁿ

=>n=1

\(A=\frac{2\cdot\left(2^3\right)^4\cdot\left(3^3\right)^8+2^2\cdot\left(2\cdot3\right)^9}{2^7\cdot6^7+2^7\cdot\left(2^3\cdot5\right).\left(3^2\right)^4}\)

\(A=\frac{2\cdot2^{12}\cdot3^{24}+2^2\cdot\left(2\cdot3\right)^9}{12^7+2^7\cdot\left(2^3\cdot5\right)\cdot3^8}\)

Đến đó thì bí

hihi

đến đó thì bt rồi

a,=1/8*1/16=1/108

b,=(27/9)^3=3^3=27

c,=125^2/25^2*25=5^2*25=25*25=225

a: \(=\dfrac{-3^4\cdot2^8}{2^2\cdot2^2\cdot3^2}=-3^2\cdot2^2=-6^2=-36\)

b: \(=\dfrac{3^6\cdot2^{15}}{3^6\cdot2^6\cdot2^{15}}=\dfrac{1}{2^6}=\dfrac{1}{64}\)

c: \(=\left(\dfrac{0.8}{0.4}\right)^5\cdot\dfrac{1}{0.4}=2^5\cdot\dfrac{1}{0.4}=\dfrac{32}{0.4}=80\)

a) \(\left(\frac{1}{3}-\frac{1}{5}\right)^2:\left(\frac{1}{5}\right)^2=\left[\left(\frac{1}{3}-\frac{1}{5}\right):\frac{1}{5}\right]^2=\left(\frac{2}{15}:\frac{1}{5}\right)^2=\left(\frac{2}{3}\right)^2=\frac{4}{9}\)

c)\(7\frac{1}{23}+\frac{10}{27}-5\frac{1}{23}+\frac{17}{27}+2^3=\left(7\frac{1}{23}-5\frac{1}{23}\right)+\left(\frac{10}{27}+\frac{17}{27}\right)+2^3=2+1+8=11\)

d)\(5.\left(-\frac{5}{2}\right)^2+\frac{1}{5}.\left(-3\right)^2=5.\frac{25}{4}+\frac{1}{5}.9=\frac{125}{4}+\frac{9}{5}=\frac{661}{20}\)

a)  \(\left(\frac{1}{3}-\frac{1}{5}\right)^2:\left(\frac{1}{5}\right)^2=\left[\left(\frac{1}{3}-\frac{1}{5}\right):\frac{1}{5}\right]^2=\left[\left(\frac{5}{15}-\frac{3}{15}\right):\frac{1}{5}\right]^2=\left[\frac{2}{15}.5\right]^2=\left[\frac{2}{3}\right]^2=\frac{4}{9}\)

b) \(\frac{2^3.5^2.8}{10^{10}}=\frac{2^3.5^2.2^3}{\left(2.5\right)^{10}}=\frac{\left(2^3.2^3\right).5^2}{2^{10}.5^{10}}=\frac{2^6.5^2}{2^{10}.5^{10}}=\frac{1}{2^4.5^8}=\frac{1}{6250000}\)

c) \(7\frac{1}{23}+\frac{10}{27}-5\frac{1}{23}+\frac{17}{27}+2^3\)

\(=\left(7\frac{1}{23}-5\frac{1}{23}\right)+\left(\frac{10}{27}+\frac{17}{27}\right)+2^3\)

\(=2+1+8\)

\(=11\)

d) \(5.\left(-\frac{5}{2}\right)^2+\frac{1}{5}.\left(-3\right)^2\)

\(=5.\frac{25}{4}+\frac{1}{5}.9\)

\(=\frac{125}{4}+\frac{9}{5}\)

\(=\frac{625}{20}+\frac{36}{20}=\frac{661}{20}\)

Chuk bạn hok tốt ! 

anh gửi chơi hay thật vậy

TẬP HỢP RA HAI NHÓM .MỘT NHÓM SỐ ÂM.CÒN NHÓM KIA LÀ SỐ DƯƠNG MÀ TÍNH

                               STUDY   WELL

      K NHA

MK XIN CẢM ƠN CÁC BẠN NHÌU

C = 24.7 −35.9 +27.10 −39.13 +...+2301.304 −3401.405 

\(C=\left(\frac{2}{4.7}+\frac{2}{7.10}+...+\frac{2}{301.304}\right)-\left(\frac{3}{5.9}+\frac{3}{9.13}+...+\frac{3}{401.405}\right)\)

\(C=\frac{2}{3}\left(\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+...+\frac{1}{301}-\frac{1}{304}\right)-\frac{3}{4}\left(\frac{1}{5}-\frac{1}{9}+\frac{1}{9}-\frac{1}{13}+...+\frac{1}{401}-\frac{1}{405}\right)\)

\(C=\frac{2}{3}\left(\frac{1}{4}-\frac{1}{304}\right)-\frac{3}{4}\left(\frac{1}{5}-\frac{1}{405}\right)\)

\(C=\frac{2}{3}.\frac{75}{304}-\frac{3}{4}.\frac{16}{81}\)

 \(C=\frac{25}{152}-\frac{4}{27}\)

\(C=\frac{67}{4104}\)

Study well 

\(C=\frac{2}{4\cdot7}-\frac{3}{5\cdot9}+\frac{2}{7\cdot10}-\frac{3}{9\cdot13}+...+\frac{2}{301\cdot304}-\frac{3}{401\cdot405}\)

\(C=\left(\frac{2}{4\cdot7}+\frac{2}{7\cdot10}+...+\frac{2}{301\cdot304}\right)-\left(\frac{3}{5\cdot9}+\frac{3}{9\cdot13}+...+\frac{3}{401\cdot405}\right)\)

\(C=\frac{2}{3}\left(\frac{3}{4\cdot7}+\frac{3}{7\cdot10}+...+\frac{3}{301\cdot304}\right)-\frac{3}{4}\left(\frac{4}{5\cdot9}+\frac{4}{9\cdot13}+...+\frac{4}{401\cdot405}\right)\)

\(C=\frac{2}{3}\left(\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+...+\frac{1}{301}-\frac{1}{304}\right)-\frac{3}{4}\left(\frac{1}{5}-\frac{1}{9}+\frac{1}{9}-\frac{1}{13}+...+\frac{1}{401}-\frac{1}{405}\right)\)

\(C=\frac{2}{3}\left(\frac{1}{4}-\frac{1}{304}\right)-\frac{3}{4}\left(\frac{1}{5}-\frac{1}{405}\right)\)

\(C=\frac{2}{3}\cdot\frac{77}{304}-\frac{3}{4}\cdot\frac{82}{405}\)

\(C=\frac{77}{456}-\frac{41}{270}\)

\(C=\frac{349}{20520}\)

Không chắc =))