Mình sửa lại để nha \(\left(125^7-625^5-25^9\right)\div99\)

Ta có \(\frac{\left(5^3\right)^7-\left(5^4\right)^6-\left(5^2\right)^9}{99}\)

\(=\frac{5^{21}-5^{20}-5^{18}}{99}\)

\(=\frac{5^{18}.\left(5^3-5^2-1\right)}{99}\)

\(=\frac{5^{18}.99}{99}\)

\(=5^{18}\)

\(\frac{25^{14}.5^{10}.625^3.125^7}{5^{14}.125^{10}.25^3.625^7}=\frac{\left(5^2\right)^{14}.5^{10}.\left(5^4\right)^3.\left(5^3\right)^7}{5^{14}.\left(5^3\right)^{10}.\left(5^2\right)^3.\left(5^4\right)^7}=\frac{5^{28}.5^{10}.5^{12}.5^{21}}{5^{14}.5^{30}.5^6.5^{28}}\)

\(=\frac{5^{71}}{5^{78}}=\frac{1}{5^7}=\frac{1}{78125}\)

\(=\dfrac{\dfrac{1}{9}-\dfrac{1}{7}-\dfrac{1}{11}}{4\left(\dfrac{1}{9}-\dfrac{1}{7}-\dfrac{1}{11}\right)}+\dfrac{3\left(\dfrac{1}{5}-\dfrac{1}{25}-\dfrac{1}{125}-\dfrac{1}{625}\right)}{4\left(\dfrac{1}{5}-\dfrac{1}{25}-\dfrac{1}{125}-\dfrac{1}{625}\right)}\)

\(=\dfrac{1}{4}+\dfrac{3}{4}=\dfrac{4}{4}=1\)

\(\dfrac{\dfrac{1}{9}-\dfrac{1}{7}-\dfrac{1}{11}}{\dfrac{4}{9}-\dfrac{4}{7}-\dfrac{4}{11}}+\dfrac{3\left(\dfrac{1}{5}-\dfrac{1}{25}-\dfrac{1}{125}-\dfrac{1}{625}\right)}{4\left(\dfrac{1}{5}-\dfrac{1}{25}-\dfrac{1}{125}-\dfrac{1}{625}\right)}\)

\(=\dfrac{\dfrac{1}{9}-\dfrac{1}{7}-\dfrac{1}{11}}{4\left(\dfrac{1}{9}-\dfrac{1}{7}-\dfrac{1}{11}\right)}+\dfrac{3}{4}\)

\(=\dfrac{1}{4}+\dfrac{3}{4}\)

\(=1\)

\(\dfrac{\dfrac{1}{9}-\dfrac{1}{7}-\dfrac{1}{11}}{\dfrac{4}{9}-\dfrac{4}{7}-\dfrac{4}{11}}+\dfrac{3\left(\dfrac{1}{5}-\dfrac{1}{25}-\dfrac{1}{125}-\dfrac{1}{625}\right)}{4\left(\dfrac{1}{5}-\dfrac{1}{25}-\dfrac{1}{125}-\dfrac{1}{625}\right)}\)

\(=\dfrac{\dfrac{1}{9}-\dfrac{1}{7}-\dfrac{1}{11}}{4\left(\dfrac{1}{9}-\dfrac{1}{7}-\dfrac{1}{11}\right)}+\dfrac{3\left(\dfrac{1}{5}-\dfrac{1}{25}-\dfrac{1}{125}-\dfrac{1}{625}\right)}{4\left(-\dfrac{1}{25}-\dfrac{1}{125}-\dfrac{1}{625}\right)}=\dfrac{1}{4}+\dfrac{3}{4}=1\)

M = 125⁷ - 625⁵ - 25⁹

M = ( 5³ )⁷ - ( 5⁴ )⁵ - ( 5² )⁹

M = 5²¹ - 5²⁰ - 5¹⁸

M = 5¹⁸ . ( 5³ - 5² - 1 )

M = 5¹⁸ . 99

M = 5¹⁸ . 9 . 11 \(⋮\)9