Ta có: \(5^2\cdot6^{11}\cdot16^2+6^2\cdot12^6\cdot15^2\)

\(=5^2\cdot2^{11}\cdot3^{11}\cdot2^8+2^2\cdot3^2\cdot2^{12}\cdot3^6\cdot5^2\cdot3^2\)

\(=2^{19}\cdot3^{11}\cdot5^2+2^{14}\cdot3^{10}\cdot5^2\)

\(=2^{14}\cdot3^{10}\cdot5^2\left(2^5\cdot3+1\right)=2^{14}\cdot3^{10}\cdot5^2\cdot97\)

Ta có: \(2\cdot6^{12}\cdot10^4-81^2\cdot960^3\)

\(=2\cdot2^{12}\cdot3^{12}\cdot2^4\cdot5^4-3^8\cdot\left(2^6\cdot3\cdot5\right)^3\)

\(=2^{17}\cdot3^{12}\cdot5^4-2^{18}\cdot3^{11}\cdot5^3\)

\(=2^{17}\cdot3^{11}\cdot5^3\left(3\cdot5-2\right)=2^{17}\cdot3^{11}\cdot5^3\cdot13\)

Ta có: \(\frac{5^2\cdot6^{11}\cdot16^2+6^2\cdot12^6\cdot15^2}{2\cdot6^{12}\cdot10^4-81^2\cdot960^3}\)

\(=\frac{2^{14}\cdot3^{10}\cdot5^2\cdot97}{2^{17}\cdot3^{11}\cdot5^3\cdot13}=\frac{97}{13\cdot2^3\cdot3\cdot5}=\frac{97}{1560}\)

\(\frac{2619}{48520}\)

đề sai

\(\frac{5^2.3^{11}.2^{11}.2^8+2^2.3^2.3^2.5^2}{2.2^{12}.3^{12}.2^4.5^4-3^8.2^{18}.3^3.5^3}\)

\(=\frac{5^2.3^{11}.2^{19}+2^2.3^4.5^2}{2^{17}.3^{12}.5^4-3^{11}.2^{18}.5^3}\)

\(=\frac{5^2.3^4.2^2.\left(3^7.2^{17}+1\right)}{2^{17}.3^{11}.5^3.\left(3.5-2\right)}=\frac{3^7.2^{17}}{2^{15}.3^7.5}+\frac{1}{2^{15}.3^7.5}=\frac{4}{5}+\frac{1}{2^{15}.3^7.5}\)

=5^2.3^11.2^11.2^8.2^2.3^2.3^6.2^12.3^2.5^2/2.3^12.2^12.2^4.5^4-3^8.3^3.2^6.5

 ban nhom cac so co cung co so o tu va mau thanh cac nhom va rut gon

=\(\frac{30\left(1+8+27+64+125\right)}{5\left(3+24+81+64+375\right)}\)

= \(\frac{30.225}{5.574}\)

=\(\frac{6750}{2870}\)

=\(\frac{675}{287}\)

K mình với!!!!

A=1.5.(3.2)+2.10.(6.2)+3.15.(9.2)+4.20.(12.2)+5.25.(15.2)

1.3.5+2.6.10+3.9.15+4.12.20+5.15.25

A=1.5.3+2.10.6+3.15.9+4.20.12+5.25.15(2.2.2.2.2)

1.3.5+2.6.10+3.9.15+4.12.20+5.15.25

A=2.2.2.2.2

A=32

\(\frac{1\cdot3\cdot5\cdot2+2\cdot10\cdot6\cdot2+3\cdot15\cdot9\cdot2+4\cdot20\cdot12\cdot2+5\cdot25\cdot15\cdot2}{1\cdot3\cdot5+2\cdot10\cdot6+3\cdot15\cdot9+4\cdot20\cdot12+5\cdot25\cdot15 }\)

\(2\cdot2\cdot2\cdot2\cdot2=2^5\)

\(=32\)

H=\(\frac{1\cdot2\cdot3+2\cdot4\cdot6+3\cdot6\cdot9+5\cdot10\cdot15}{1\cdot3\cdot6+2\cdot6\cdot12+3\cdot9\cdot18+5\cdot15\cdot30}=\frac{1.2.3+2^3.\left(1.2.3\right)+3^3.\left(1.2.3\right)+5^3.\left(1.2.3\right)}{1.3.6+2^3.\left(1.3.6\right)+3^3.\left(1.3.6\right)+5^3.\left(1.3.6\right)}=\frac{1.2.3.\left(1+2^3+3^3+5^3\right)}{1.3.6.\left(1+2^3+3^3+5^3\right)}=\frac{2}{6}=\frac{1}{3}\)

\(\frac{4}{3}\)