a) \(204-84:12=204-7=197\)

b) \(15.2^3+4.3^2-5.7=15.8+4.9+5.7=120+36+35=156+35=191\)

c) \(5^6:5^3+2^3.2^2=5^3+2^5=125+32=157\)

d) \(164.53+47.164=164.\left(53+47\right)=164.100=16400\)

_Chúc bạn học tốt_

Để A thì 

\(A=\left(5m^2-8m^2-9m^2\right)\left(-n^3+4n^3\right)=-12m^2.3n^3=-36m^2n^3\)

Để A\(\ge0\) thì \(m^2n^3\le0\)

\(\Leftrightarrow\left\{{}\begin{matrix}m\in Q\\n\le0\end{matrix}\right.\)

Bằng nhau

số cuối là mấy vậy bạn

a) \(\left(\frac{-1}{3}\right)^4=\frac{\left(-1\right)^4}{3^4}=\frac{1}{81}\)

b) \(\left(-2\frac{1}{4}\right)^3=\left(\frac{-9}{4}\right)^3=\frac{\left(-9\right)^3}{4^3}=\frac{-729}{64}\)

c) \(\left(-0,2\right)^2=\left(\frac{-1}{5}\right)^2=\frac{\left(-1\right)^2}{5^2}=\frac{1}{25}\)

d) \(\left(-5,3\right)^0=1\)

a)\(\left(\frac{-1}{3}\right)^4=\frac{1}{81}\)

b) \(\left(-2\frac{1}{4}\right)^3=\frac{-729}{64}\)

c) \(\left(-0,2\right)^2=\frac{1}{25}\)

d) \(\left(-5,3\right)^0=1\)

Cbht

a) 1/81

b) 9261/64

c)0.04

d)0

1)Tính:

a)\(4^2\cdot2=\left(2^2\right)^2\cdot2=2^4\cdot2=2^5=32\)

b)\(36^2:6^2=\left(36:6\right)^2=6^2=48\)

c)\(\left(\frac{2}{5}\right)^{10}:\left(\frac{4}{25}\right)^2=\left(\frac{2}{5}\right)^{10}\cdot\left(\frac{25}{4}\right)^2=\)\(\left(1\right)^{10}\cdot\left(\frac{5}{2}\right)^2=1\cdot\frac{5^2}{2^2}=1\cdot\frac{25}{4}=\frac{25}{4}\)

a

\(4^2.2=16.2=32\)

b\(36^2:6^2=36.36:6.6=36.36:36=36\)

c

3)

a)\(x:\left(-\frac{1}{3}\right)^5=\left(-\frac{1}{3}\right)^3\)

                      \(x=\left(-\frac{1}{3}\right)^3\cdot\left(-\frac{1}{3}\right)^5\)           

                      \(x=\left(-\frac{1}{3}\right)^8\)

b)\(\left(\frac{2}{3}\right)^{16}\cdot x=\left(\frac{2}{3}\right)^{18}\)

                      \(x=\left(\frac{2}{3}\right)^{18}:\left(\frac{2}{3}\right)^{16}\)

                      \(x=\left(\frac{2}{3}\right)^2\)

                      \(x=\frac{2^2}{3^2}\)

                      \(x=\frac{4}{9}\)

`#3107.101107`

\(A = 2 + 2^2 + 2^3 + ... + 2^{2020} + 2^{2021} + 2^{2022}\)

\(= (2 + 2^2) + (2^3 + 2^4) + ... + (2^{2021} + 2^{2022})\)

\(=2(1+2) + 2^3(1 + 2) + ... + 2^{2021}(1 + 2)\)

\(=(1 + 2)(2 + 2^3 + ... + 2^{2021})\)

\(= 3(2 + 2^3 + ... + 2^{2021})\)

Vì \(3(2 + 2^3 + ... + 2^{2021})\) \(\vdots\) \(3\)

`\Rightarrow A \vdots 3`

Vậy, `A \vdots 3.`