\(\left(2^5\cdot9-2^5\cdot2018\right):2^7\)

\(=2^5\left(9-2018\right):2^7\)

\(=2^5\cdot\left(-2009\right):2^7\)

\(=-2009:2^2\)

\(=-2009:4\)

\(=\frac{-2009}{4}\)

3 chữ số đó là 721, 631, 532

721;631;532

\(\left(13.3^3+14.3^3-15.3^3\right):3^4\)

\(=[3^3\left(13+14-15\right)]:3^4\)

\(=\left(3^3.12\right):3^4\)

\(=\left(27.12\right):81\)

\(=324:81=4\)

\(\left(13\cdot3^3+14\cdot3^3-15\cdot3^3\right):3^4\)

\(=3^3\left(13+14-15\right):3^4\)

\(=3^3\cdot12:3^4\)

\(=3^4\cdot4:3^4\)

\(=4\)

Ta có: 

\(2^{100}=\left(2^{10}\right)^{10}=1024^{10}\)

Vì: \(1024^{10}>1024^9\)

nên \(2^{100}>1024^9\)

so sánh 2¹⁰⁰ và 1024⁹

giải 

ta có:

1024=32^2.9=32¹⁸=2¹⁸.16¹⁸

^{ta có}:16¹⁸=4^2.18=4³⁶=2^2.36=2⁷²

^{ta có:}2¹⁸.2⁷²=2⁹⁰

mà 2⁹⁰<2¹⁰⁰

=>2¹⁰⁰ > 1024⁹

Ta có:

A=abc + bca + cab

=>A=(100a+10b+c)+(100b+10c+a)+(100c+10a+b)

=>A=100a + 10b + c + 100b + 10c + a + 100c + 10a + b

=>A=111a + 111b + 111c

=>A=111(a+b+c)

giúp mik với mik sắp phải nộp bài rùi

Số đó là 368

3A=\(3+3^2+3^2+...+3^{51}\)
\(3A-A=3^{51}-1\)

\(2A=3^{51}-1\)

\(2A+1=3^{51}\)
Ta có: \(3^{51}+1=3^n+1\Leftrightarrow3^{51}=3^n\Leftrightarrow n=51\)

Ta có:

\(3^{11}< 32^{11}=\left(4.8\right)^{11}=4^{11}.8^{11}=2^{22}.8^{11}\)

\(17^{14}>16^{14}=2^{14}.8^{14}=2^{14}.8^3.8^{11}=2^{14}.2^9.8^{11}=2^{23}.8^{11}\)

Vì \(2^{33}.8^{11}>2^{22}.8^{11}\Rightarrow16^{14}>32^{11}\)

Vậy \(17^{14}>16^{14}>32^{11}>31^{11}\Rightarrow17^{14}>31^{11}\)

12 ml = 12cc

12ml=12cc đúng rồi còn gì