\(A=\frac{4^5.9^4-2.6^9}{2^{10}.3^8-6^8.20}\)

\(A=\frac{\left(2^2\right)^5.\left(3^2\right)^4-2.\left(2.3\right)^9}{2^{10}.3^8-\left(2.3\right)^8.2^2.5}\)

\(A=\frac{2^{10}.3^8-2^{10}.3^9}{2^{10}.3^8-2^{10}.3^8.5}\)

\(A=\frac{2^{10}.\left(3^8-3^9\right)}{2^{10}.3^8.\left(1-5\right)}=\frac{3^8-3^9}{3^8.\left(-4\right)}=\frac{3^8.\left(1-3\right)}{3^8.\left(-4\right)}=\frac{-2}{-4}=\frac{1}{2}\)

Vậy A = \(\frac{1}{2}\)

\(B=\frac{2^{19}.27^3+15.4^9.9^4}{6^9.2^{10}+12^{10}}\)

\(B=\frac{2^{19}.\left(3^3\right)^3+3.5.\left(2^2\right)^9.\left(3^2\right)^4}{\left(2.3\right)^9.2^{10}+\left(2^2.3\right)^{10}}\)

\(B=\frac{2^{19}.3^9+3.5.2^{18}.3^8}{2^9.3^9.2^{10}+2^{20}.3^{10}}\)

\(B=\frac{2^{19}.3^9+3^9.2^{18}.5}{2^{19}.3^9+2^{20}.3^{10}}\)

\(B=\frac{2^{18}.3^9.\left(2+5\right)}{2^{19}.3^9\left(1+2.3\right)}=\frac{7}{2.7}=\frac{1}{2}\)

Vậy B = \(\frac{1}{2}\)

ban nao co chuyen shin ko cho minh muon minh giai cho 10 bai nhu the i love pac pac

\(A=\frac{2^{12}.3^4-4^5.9^2}{\left(2^2.3\right)^6+8^4.3^5}\)

\(A=\frac{2^{12}.3^4-2^{10}.3^4}{2^{12}.3^6+2^{12}.3^5}\)

\(A=\frac{2^{10}.3^4\left(2^2-1\right)}{2^{10}.3^4\left(2^2.3^2+2^2.3\right)}\)

\(A=\frac{2^2-1}{2^2.3^2+2^2.3}\)

\(A=\frac{4-1}{36+12}\)

\(A=\frac{3}{48}=\frac{1}{16}\)

a: \(=\dfrac{2^4\cdot3^6\cdot2\cdot3}{2^4\cdot3^6}=6\)

b: \(=\dfrac{2^{20}\cdot3^{20}}{2^{18}\cdot3^{18}}=2^2\cdot3^2=36\)

c: \(=\dfrac{12^5\cdot13}{12^6\cdot13}-\dfrac{12^8\cdot\left(-11\right)}{12^9\cdot\left(-11\right)}=\dfrac{1}{12}-\dfrac{1}{12}=0\)

a) \(A=\frac{2^4.25^4}{10^5.5^5}=\frac{2^4.\left(5^2\right)^4}{\left(2.5\right)^5.5^5}=\frac{2^4.5^8}{2^5.5^{10}}=\frac{1}{2.5^2}=\frac{1}{50}\)

b) \(B=\frac{4^5.9^4-2.6^9}{2^{10}.3^8+6^8.20}\)

\(B=\frac{2^{10}.3^8-2^{10}.3^9}{2^{10}.3^8+2^{10}.3^8.5}=\frac{2^{10}.3^8.\left(1-3\right)}{2^{10}.3^8.\left(1+5\right)}=\frac{-2}{6}=\frac{-1}{3}\)

\(A=\frac{2^4.25^4}{10^5.5^5}\)

\(A=\frac{2^4.\left(5^2\right)^4}{\left(2.5\right)^5.5^5}\)

\(A=\frac{2^4.5^8}{2^5.5^5.5^5}\)

\(A=\frac{5^8}{2.5^{10}}\)

\(A=\frac{1}{2.5^2}=\frac{1}{2.25}=\frac{1}{50}\)

\(B=\frac{4^5.9^4-2.6^9}{2^{10}.3^8+6^8.20}\)

\(B=\frac{\left(2^2\right)^5.\left(3^2\right)^4-2.\left(2.3\right)^9}{2^{10}.3^8+\left(2.3\right)^8.2^2.5}\)

\(B=\frac{2^{10}.3^8-2^{10}.2.3^9}{2^{10}.3^8+2^{10}.3^8.5}\)

\(B=\frac{2^{10}.\left(3^8-2.3^9\right)}{2^{10}.\left(3^8+3^8.5\right)}\)

\(\Rightarrow B=\frac{3^8.2.3^9}{3^8+3^8.5}\) ( \(2^{10}\ne0\))

\(B=\frac{3^8.\left(1-2.3\right)}{3^8.\left(1+5\right)}\)

\(B=\frac{1-6}{1+5}\left(3^8\ne0\right)\)

\(B=\frac{-5}{6}\)

Tham khảo nhé~

1/

a, 24 - {12 -[(-10)-(-2)]}

=24-{12-(-8)}

=24-20

=4

b,\(\frac{2}{3}\)+\(\frac{1}{3}\).(\(\frac{-2}{3}\)+\(\frac{5}{6}\)):\(\frac{2}{3}\)

=\(\frac{2}{3}\)+\(\frac{1}{3}\).(\(\frac{-4}{6}\)+\(\frac{5}{6}\)):\(\frac{2}{3}\)

=\(\frac{2}{3}\)+\(\frac{1}{3}\).\(\frac{1}{6}\).\(\frac{3}{2}\)

=\(\frac{2}{3}\)+\(\frac{1}{18}\).\(\frac{3}{2}\)

=\(\frac{2}{3}\)+\(\frac{1}{6}\).\(\frac{1}{2}\)

=\(\frac{2}{3}\)+\(\frac{1}{12}\)

=\(\frac{8}{12}\)+\(\frac{1}{12}\)

=\(\frac{9}{12}\)=\(\frac{3}{4}\)

2/

a,\(\frac{-2}{5}\)+(\(\frac{-5}{9}\)+\(\frac{2}{5}\))

=\(\frac{-2}{5}\)+\(\frac{-5}{9}\)+\(\frac{2}{5}\)

=(\(\frac{-2}{5}\)+\(\frac{2}{5}\))+\(\frac{-5}{9}\)

=         0+\(\frac{-5}{9}\)

=          \(\frac{-5}{9}\)

\(1,\)

\(\dfrac{45^2.3^8.10^5}{5^5.3^7.18^5}\)

\(=\dfrac{3^4.5^2.3^8.2^5.5^5}{5^5.3^7.2^5.3^{10}}\)

\(=\dfrac{3^{12}.2^5.5^7}{5^5.3^{17}.2^5}\)

\(=\dfrac{1.5^2}{3^5.1}\)

\(=\dfrac{25}{243}\)

\(2,\)

\(\dfrac{4^5.9^4+2.6^9}{2^{10}.3^8+6^8.20}\)

\(=\dfrac{2^{10}.3^8+2.2^9.3^9}{2^{10}.3^8+2^8.3^8.2^2.5}\)

\(=\dfrac{2^{10}.3^8+2^{10}.3^9}{2^{10}.3^8+2^{10}.3^8.5}\)

\(=\dfrac{2^{10}.3^8.4}{2^{10}.3^8.6}\)

\(=\dfrac{2^{12}.3^8}{2^{11}.3^9}\)

\(=\dfrac{2}{3}\)

\(3,\)

\(\dfrac{15.3^{11}+4.27^4}{9^7}\)

\(=\dfrac{3.5.3^{11}+2^2.3^{12}}{3^{14}}\)

\(=\dfrac{5.3^{12}+2^2.3^{12}}{3^{14}}\)

\(=\dfrac{3^{12}\left(5+2^2\right)}{3^{14}}\)

\(=\dfrac{3^{12}.9}{3^{14}}\)

\(=\dfrac{3^{14}}{3^{14}}\)

\(=1\)

\(4,\)

\(\dfrac{4^7.2^8}{3.2^{15}.16^2-5^2\left(2^{10}\right)^2}\)

\(=\dfrac{2^{22}}{3.2^{23}-5^2.2^{20}}\)

\(=\dfrac{2^{22}}{2^{20}.\left(-1\right)}\)

\(=\dfrac{2^{22}}{-2^{20}}\)

\(=-4\)

* Mấy bài còn lại tương tự đấy bạn tự làm đi

Mình mỏi tay lắm rồi

P/s:khuyến khích tự làm,chỉ làm mẫu 1 câu:

1)\(\dfrac{45^2.3^8.10^5}{5^5.3^7.18^5}=\dfrac{\left(5.9\right)^2.3.3^7.\left(2.5\right)^5}{5^5.3^7.\left(2.9\right)^5}\)\(=\dfrac{5^2.9^2.3.3^7.2^5.5^5}{5^5.3^7.2^5.9^5}\)\(=\dfrac{5^2.9^2.3.1.1.1}{1.1.1.9^5}\)\(=\dfrac{5^2.9^2.3}{9^5}=\dfrac{5^2.9^2.3}{9^2.9^3}=\dfrac{5^2.3}{9^3}=\dfrac{75}{729}=\dfrac{25}{243}\)

A = \(\frac{2}{3}+\frac{5}{6}:6-\frac{1}{8}.\left(-3\right)^2\)

A = \(\frac{2}{3}+\frac{5}{6}.\frac{1}{6}-\frac{1}{8}.9\)

A = \(\frac{2}{3}+\frac{5}{36}-\frac{9}{8}\)

A =\(\frac{-23}{72}\)

\(C=\frac{2^{19}.3^9+5.2^{18}.3^9}{2^9.3^9.2^{10}+3^{10}.2^{20}}\)

\(C=\frac{3^9.\left(2^{19}+5.2^{18}\right)}{3^9.\left(2^9.2^{10}+3.2^{20}\right)}\)

\(C=\frac{2^{19}+5.2^{18}}{2^9.2^{10}+3.2^{20}}\)suy ra \(C=\frac{2^{18}.\left(2+5\right)}{2^{18}.\left(2+3.2^2\right)}\)

\(C=\frac{7}{14}\)

Vậy \(C=\frac{1}{2}\)

treu ti thoi

problems?